From 29e04bd4f13730eadbe6b1c20529f8af807119f3 Mon Sep 17 00:00:00 2001 From: zhangzhen Date: Thu, 16 Jul 2020 17:35:27 +0800 Subject: [PATCH] =?UTF-8?q?add:=E5=A2=9E=E5=8A=A0gitee=E9=95=9C=E5=83=8F?= =?UTF-8?q?=E6=BA=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- scripts/constants.js | 16 ++- src/db/root.db.js | 305 +++++++++++++++++++++++++++++++++++++++---- 2 files changed, 288 insertions(+), 33 deletions(-) diff --git a/scripts/constants.js b/scripts/constants.js index e62a081..65f1bfe 100644 --- a/scripts/constants.js +++ b/scripts/constants.js @@ -1,10 +1,13 @@ + +const CRAWL_SITE = "gitee" + module.exports = { /** * 爬取的数据源站点 github | gitee */ - CRAWL_SITE: "github", + CRAWL_SITE, /** * 需解析的语言类型 @@ -35,18 +38,19 @@ module.exports = { /** * 获取问题列表地址 */ - PROBLEMS_URL: "https://github.com/azl397985856/leetcode/tree/master/problems", + PROBLEMS_URL: CRAWL_SITE === "gitee" ? "https://gitee.com/golong/leetcode/tree/master/problems" : "https://github.com/azl397985856/leetcode/tree/master/problems", /** * 抓取页面问题内容的dom元素选择器 */ - QUESTION_DOM_SELECTOR: ".position-relative.js-navigation-item .js-navigation-open", + QUESTION_DOM_SELECTOR: CRAWL_SITE === "gitee" ? ".tree-item-file-name.tree-list-item a" : ".position-relative.js-navigation-item .js-navigation-open", + /** * markdwon下载地址 */ - BASE_MARKDWON_DOWNLOAD_URL: - "https://raw.githubusercontent.com/azl397985856/leetcode/master/problems/", + BASE_MARKDWON_DOWNLOAD_URL: CRAWL_SITE === "gitee" ? "https://gitee.com/golong/leetcode/raw/master/problems/" : "https://raw.githubusercontent.com/azl397985856/leetcode/master/problems/", + /** * 过滤英文文档末尾标识 @@ -54,7 +58,7 @@ module.exports = { ENGLISH_MARKDOWN_SIGN: ".en.md", /** - * 爬虫抓取同一文件时的最大失败次数(多为网络原因导致) + * 爬虫抓取同一文件时的最大失败次数(多为网络原因导致, 切换到gitee源) */ MAX_CRAWL_RETRY_NUMBER : 100 }; diff --git a/src/db/root.db.js b/src/db/root.db.js index a6e95a8..2d89f2c 100644 --- a/src/db/root.db.js +++ b/src/db/root.db.js @@ -1525,7 +1525,7 @@ "code": [ { "language": "java", - "text": "\nclass Solution {\n public int findTheLongestSubstring(String s) {\n\n int len = s.length();\n\n if (len == 0)\n return 0;\n\n int[][] preSum = new int[len][5];\n int start = getIndex(s.charAt(0));\n if (start != -1)\n preSum[0][start]++;\n\n // preSum\n for (int i = 1; i < len; i++) {\n\n int idx = getIndex(s.charAt(i));\n\n for (int j = 0; j < 5; j++) {\n\n if (idx == j)\n preSum[i][j] = preSum[i - 1][j] + 1;\n else\n preSum[i][j] = preSum[i - 1][j];\n }\n }\n\n for (int i = len - 1; i >= 0; i--) {\n\n for (int j = 0; j < len - i; j++) {\n if (checkValid(preSum, s, i, i + j))\n return i + 1\n }\n }\n return 0\n }\n\n\n public boolean checkValid(int[][] preSum, String s, int left, int right) {\n\n int idx = getIndex(s.charAt(left));\n\n for (int i = 0; i < 5; i++)\n if (((preSum[right][i] - preSum[left][i] + (idx == i ? 1 : 0)) & 1) == 1)\n return false;\n\n return true;\n }\n public int getIndex(char ch) {\n\n if (ch == 'a')\n return 0;\n else if (ch == 'e')\n return 1;\n else if (ch == 'i')\n return 2;\n else if (ch == 'o')\n return 3;\n else if (ch == 'u')\n return 4;\n else\n return -1;\n }\n}\n" + "text": "\nclass Solution {\n public int findTheLongestSubstring(String s) {\n\n int len = s.length();\n\n if (len == 0)\n return 0;\n\n int[][] preSum = new int[len][5];\n int start = getIndex(s.charAt(0));\n if (start != -1)\n preSum[0][start]++;\n\n // preSum\n for (int i = 1; i < len; i++) {\n\n int idx = getIndex(s.charAt(i));\n\n for (int j = 0; j < 5; j++) {\n\n if (idx == j)\n preSum[i][j] = preSum[i - 1][j] + 1;\n else\n preSum[i][j] = preSum[i - 1][j];\n }\n }\n\n for (int i = len - 1; i >= 0; i--) {\n\n for (int j = 0; j < len - i; j++) {\n if (checkValid(preSum, s, j, i + j))\n return i + 1;\n }\n }\n return 0;\n }\n\n\n public boolean checkValid(int[][] preSum, String s, int left, int right) {\n\n int idx = getIndex(s.charAt(left));\n\n for (int i = 0; i < 5; i++)\n if (((preSum[right][i] - preSum[left][i] + (idx == i ? 1 : 0)) & 1) == 1)\n return false;\n\n return true;\n }\n public int getIndex(char ch) {\n\n if (ch == 'a')\n return 0;\n else if (ch == 'e')\n return 1;\n else if (ch == 'i')\n return 2;\n else if (ch == 'o')\n return 3;\n else if (ch == 'u')\n return 4;\n else\n return -1;\n }\n}\n" }, { "language": "py", @@ -2326,6 +2326,22 @@ { "language": "js", "text": "\n/**\n * @param {string} s\n * @return {boolean}\n */\nvar isValid = function (s) {\n let valid = true;\n const stack = [];\n const mapper = {\n \"{\": \"}\",\n \"[\": \"]\",\n \"(\": \")\",\n };\n\n for (let i in s) {\n const v = s[i];\n if ([\"(\", \"[\", \"{\"].indexOf(v) > -1) {\n stack.push(v);\n } else {\n const peak = stack.pop();\n if (v !== mapper[peak]) {\n return false;\n }\n }\n }\n\n if (stack.length > 0) return false;\n\n return valid;\n};\n" + }, + { + "language": "js", + "text": "\nvar isValid = function (s) {\n while (s.includes(\"[]\") || s.includes(\"()\") || s.includes(\"{}\")) {\n s = s.replace(\"[]\", \"\").replace(\"()\", \"\").replace(\"{}\", \"\");\n }\n s = s.replace(\"[]\", \"\").replace(\"()\", \"\").replace(\"{}\", \"\");\n return s.length === 0;\n};\n" + }, + { + "language": "cpp", + "text": "\nclass Solution {\npublic:\n bool isValid(string s) {\n int top = -1;\n for(int i =0;i 0 && target < matrix[rowIndex][colIndex]) {\n rowIndex --;\n }\n\n while(colIndex < matrix[0].length) {\n if (target === matrix[rowIndex][colIndex]) return true;\n if (target > matrix[rowIndex][colIndex]) {\n colIndex ++;\n } else if (rowIndex > 0){\n rowIndex --;\n } else {\n return false;\n }\n }\n\n return false;\n};\n" + "text": "\n\n/*\n * @lc app=leetcode id=240 lang=javascript\n *\n * [240] Search a 2D Matrix II\n *\n * https://leetcode.com/problems/search-a-2d-matrix-ii/description/\n *\n * \n */\n/**\n * @param {number[][]} matrix\n * @param {number} target\n * @return {boolean}\n */\nvar searchMatrix = function(matrix, target) {\n if (!matrix || matrix.length === 0) return false;\n\n let colIndex = 0;\n let rowIndex = matrix.length - 1;\n while(rowIndex > 0 && target < matrix[rowIndex][colIndex]) {\n rowIndex --;\n }\n\n while(colIndex < matrix[0].length) {\n if (target === matrix[rowIndex][colIndex]) return true;\n if (target > matrix[rowIndex][colIndex]) {\n colIndex ++;\n } else if (rowIndex > 0){\n rowIndex --;\n } else {\n return false;\n }\n }\n\n return false;\n};\n" }, { "language": "py", @@ -3434,11 +3445,15 @@ }, { "language": "js", - "text": "\n/**\n * @param {ListNode} head\n * @param {number} k\n * @return {ListNode}\n */\nvar reverseKGroup = function(head, k) {\n // 标兵\n let dummy = new ListNode()\n dummy.next = head\n let [start, end] = [dummy, dummy.next]\n let count = 0\n while(end) {\n count++\n if (count % k === 0) {\n start = reverseList(start, end.next)\n end = start.next\n } else {\n end = end.next\n }\n }\n return dummy.next\n\n // 翻转stat -> end的链表\n function reverseList(start, end) {\n let [pre, cur] = [start, start.next]\n const first = cur\n while(cur !== end) {\n let next = cur.next\n cur.next = pre\n pre = cur\n cur = next\n }\n start.next = pre\n first.next = cur\n return first\n }\n};\n\n" + "text": "\n/**\n * @param {ListNode} head\n * @param {number} k\n * @return {ListNode}\n */\nvar reverseKGroup = function (head, k) {\n // 标兵\n let dummy = new ListNode();\n dummy.next = head;\n let [start, end] = [dummy, dummy.next];\n let count = 0;\n while (end) {\n count++;\n if (count % k === 0) {\n start = reverseList(start, end.next);\n end = start.next;\n } else {\n end = end.next;\n }\n }\n return dummy.next;\n\n // 翻转stat -> end的链表\n function reverseList(start, end) {\n let [pre, cur] = [start, start.next];\n const first = cur;\n while (cur !== end) {\n let next = cur.next;\n cur.next = pre;\n pre = cur;\n cur = next;\n }\n start.next = pre;\n first.next = cur;\n return first;\n }\n};\n" }, { "language": "py", - "text": "\nclass Solution:\n def reverseKGroup(self, head: ListNode, k: int) -> ListNode:\n if head is None or k < 2:\n return head\n dummy = ListNode(0)\n dummy.next = head\n start = dummy\n end = head\n count = 0\n while end:\n count += 1\n if count % k == 0:\n start = self.reverse(start, end.next)\n end = start.next\n else:\n end = end.next\n return dummy.next\n\n def reverse(self, start, end):\n prev, curr = start, start.next\n first = curr\n while curr != end:\n temp = curr.next\n curr.next = prev\n prev = curr\n curr = temp\n start.next = prev\n first.next = curr\n return first\n" + "text": "\nclass Solution:\n def reverseKGroup(self, head: ListNode, k: int) -> ListNode:\n if head is None or k < 2:\n return head\n dummy = ListNode(0)\n dummy.next = head\n start = dummy\n end = head\n count = 0\n while end:\n count += 1\n if count % k == 0:\n start = self.reverse(start, end.next)\n # end 调到下一个\n end = start.next\n else:\n end = end.next\n return dummy.next\n # (start, end） 左右都开放\n\n def reverse(self, start, end):\n prev, curr = start, start.next\n first = curr\n # 反转\n while curr != end:\n next = curr.next\n curr.next = prev\n prev = curr\n curr = next\n # 将反转后的链表添加到原链表中\n start.next = prev\n first.next = end\n # 返回反转前的头， 也就是反转后的尾部\n return first\n\n" + }, + { + "language": "py", + "text": "\n\nclass Solution:\n def reverseKGroup(self, head: ListNode, k: int) -> ListNode:\n if head is None or k < 2:\n return head\n dummy = ListNode(0)\n dummy.next = head\n pre = dummy\n cur = head\n count = 0\n while cur:\n count += 1\n if count % k == 0:\n pre = self.reverse(pre, cur.next)\n # end 调到下一个位置\n cur = pre.next\n else:\n cur = cur.next\n return dummy.next\n # (p1, p4） 左右都开放\n\n def reverse(self, p1, p4):\n prev, curr = p1, p1.next\n p2 = curr\n # 反转\n while curr != p4:\n next = curr.next\n curr.next = prev\n prev = curr\n curr = next\n # 将反转后的链表添加到原链表中\n # prev 相当于 p3\n p1.next = prev\n p2.next = p4\n # 返回反转前的头， 也就是反转后的尾部\n return p2\n\n# @lc code=end\n\n" } ] }, @@ -4044,7 +4059,7 @@ "color": "blue" }, { - "text": "1],因此,需要检查越界;3.第4点特征最容易遗漏,还有就是不需要检查越界,因为根据定义可知:i>=dp[i],所以dp[i", + "text": "1],因此,需要检查越界;2.第4点特征最容易遗漏,还有就是不需要检查越界,因为根据定义可知:i>=dp[i],所以dp[i", "link": null, "color": "blue" }, @@ -4056,13 +4071,33 @@ ], "solution": "https://github.com/azl397985856/leetcode/blob/master/problems/32.longest-valid-parentheses.md", "code": [ + { + "language": "java", + "text": "\npublic class Solution {\n public int longestValidParentheses(String s) {\n int left = 0, right = 0, maxlength = 0;\n for (int i = 0; i < s.length(); i++) {\n if (s.charAt(i) == '(') {\n left++;\n } else {\n right++;\n }\n if (left == right) {\n maxlength = Math.max(maxlength, left + right);\n }\n if (right > left) {\n left = right = 0;\n }\n }\n left = right = 0;\n for (int i = s.length() - 1; i >= 0; i--) {\n if (s.charAt(i) == '(') {\n left++;\n } else {\n right++;\n }\n if (left == right) {\n maxlength = Math.max(maxlength, left + right);\n }\n if (left > right) {\n left = right = 0;\n }\n }\n return maxlength;\n }\n}\n" + }, { "language": "js", - "text": "\n// 用栈来解\nvar longestValidParentheses = function(s) {\n let stack = new Array()\n let longest = 0\n stack.push(-1)\n for(let i = 0; i < s.length; i++) {\n if (s[i] === '(') {\n stack.push(i)\n } else {\n stack.pop()\n if (stack.length === 0) {\n stack.push(i)\n } else {\n longest = Math.max(longest, i - stack[stack.length - 1])\n }\n }\n }\n return longest\n};\n" + "text": "\n// 用栈来解\nvar longestValidParentheses = function (s) {\n let stack = new Array();\n let longest = 0;\n stack.push(-1);\n for (let i = 0; i < s.length; i++) {\n if (s[i] === \"(\") {\n stack.push(i);\n } else {\n stack.pop();\n if (stack.length === 0) {\n stack.push(i);\n } else {\n longest = Math.max(longest, i - stack[stack.length - 1]);\n }\n }\n }\n return longest;\n};\n" + }, + { + "language": "py", + "text": "\nclass Solution:\n def longestValidParentheses(self, s: str) -> int:\n n = len(s)\n ans = 0\n\n def validCnt(start):\n # cnt 为 ) 的数量减去 ( 的数量\n cnt = 0\n ans = 0\n for i in range(start, n):\n if s[i] == '(':\n cnt += 1\n if s[i] == ')':\n cnt -= 1\n if cnt < 0:\n return i - start\n if cnt == 0:\n ans = max(ans, i - start + 1)\n return ans\n for i in range(n):\n ans = max(ans, validCnt(i))\n\n return ans\n" + }, + { + "language": "py", + "text": "\n\nclass Solution:\n def longestValidParentheses(self, s: str) -> int:\n if not s:\n return 0\n res = 0\n stack = [-1]\n for i in range(len(s)):\n if s[i] == \"(\":\n stack.append(i)\n else:\n stack.pop()\n if not stack:\n stack.append(i)\n else:\n res = max(res, i - stack[-1])\n return res\n" + }, + { + "language": "py", + "text": "\nclass Solution:\n def longestValidParentheses(self, s: str) -> int:\n ans = l = r = 0\n for c in s:\n if c == '(':\n l += 1\n else:\n r += 1\n if l == r:\n ans = max(ans, l + r)\n if r > l:\n l = r = 0\n l = r = 0\n for c in s[::-1]:\n if c == '(':\n l += 1\n else:\n r += 1\n if l == r:\n ans = max(ans, l + r)\n if r < l:\n l = r = 0\n\n return ans\n" }, { "language": "py", "text": "\ns = '(())())'\n" + }, + { + "language": "py", + "text": "\nclass Solution:\n def longestValidParentheses(self, s: str) -> int:\n mlen = 0\n slen = len(s)\n dp = [0] * (slen + 1)\n for i in range(1, len(s) + 1):\n # 有效的括号对不可能会以'('结尾的\n if s[i - 1] == '(':\n continue\n\n left_paren = i - 2 - dp[i - 1]\n if left_paren >= 0 and s[left_paren] == '(':\n dp[i] = dp[i - 1] + 2\n\n # 拼接有效括号对\n if dp[i - dp[i]]:\n dp[i] += dp[i - dp[i]]\n\n # 更新最大有效扩对长度\n if dp[i] > mlen:\n mlen = dp[i]\n\n return mlen\n" } ] }, @@ -4756,7 +4791,23 @@ "id": "42", "name": "trapping-rain-water", "company": [], - "pre": [], + "pre": [ + { + "text": " 空间换时间", + "link": null, + "color": "red" + }, + { + "text": " 双指针", + "link": null, + "color": "red" + }, + { + "text": " 单调栈", + "link": null, + "color": "red" + } + ], "keyPoints": [ { "text": "建模`h[i]=Math.min(左边柱子最大值,右边柱子最大值)`(h为下雨之后的水位)", @@ -4768,15 +4819,27 @@ "code": [ { "language": "js", - "text": "\n\nfor(let i = 0; i < height.length; i++) {\n area += (h[i] - height[i]) * 1; // h为下雨之后的水位\n}\n\n" + "text": "\nfor (let i = 0; i < height.length; i++) {\n area += (h[i] - height[i]) * 1; // h为下雨之后的水位\n}\n" }, { "language": "js", - "text": "\n\n/*\n * @lc app=leetcode id=42 lang=javascript\n *\n * [42] Trapping Rain Water\n *\n */\n/**\n * @param {number[]} height\n * @return {number}\n */\nvar trap = function(height) {\n let max = 0;\n let volumn = 0;\n const leftMax = [];\n const rightMax = [];\n\n for(let i = 0; i < height.length; i++) {\n leftMax[i] = max = Math.max(height[i], max);\n }\n\n max = 0;\n\n for(let i = height.length - 1; i >= 0; i--) {\n rightMax[i] = max = Math.max(height[i], max);\n }\n\n for(let i = 0; i < height.length; i++) {\n volumn = volumn + Math.min(leftMax[i], rightMax[i]) - height[i]\n }\n\n return volumn;\n};\n\n" + "text": "\n/*\n * @lc app=leetcode id=42 lang=javascript\n *\n * [42] Trapping Rain Water\n *\n */\n/**\n * @param {number[]} height\n * @return {number}\n */\nvar trap = function (height) {\n let max = 0;\n let volume = 0;\n const leftMax = [];\n const rightMax = [];\n\n for (let i = 0; i < height.length; i++) {\n leftMax[i] = max = Math.max(height[i], max);\n }\n\n max = 0;\n\n for (let i = height.length - 1; i >= 0; i--) {\n rightMax[i] = max = Math.max(height[i], max);\n }\n\n for (let i = 0; i < height.length; i++) {\n volume = volume + Math.min(leftMax[i], rightMax[i]) - height[i];\n }\n\n return volume;\n};\n" + }, + { + "language": "cpp", + "text": "\nint trap(vector& heights)\n{\n\tif(heights == null)\n\t\treturn 0;\n int ans = 0;\n int size = heights.size();\n vector left_max(size), right_max(size);\n left_max[0] = heights[0];\n for (int i = 1; i < size; i++) {\n left_max[i] = max(heights[i], left_max[i - 1]);\n }\n right_max[size - 1] = heights[size - 1];\n for (int i = size - 2; i >= 0; i--) {\n right_max[i] = max(heights[i], right_max[i + 1]);\n }\n for (int i = 1; i < size - 1; i++) {\n ans += min(left_max[i], right_max[i]) - heights[i];\n }\n return ans;\n}\n\n" + }, + { + "language": "cpp", + "text": "\n\nclass Solution {\npublic:\n int trap(vector& heights)\n{\n int left = 0, right = heights.size() - 1;\n int ans = 0;\n int left_max = 0, right_max = 0;\n while (left < right) {\n if (heights[left] < heights[right]) {\n heights[left] >= left_max ? (left_max = heights[left]) : ans += (left_max - heights[left]);\n ++left;\n }\n else {\n heights[right] >= right_max ? (right_max = heights[right]) : ans += (right_max - heights[right]);\n --right;\n }\n }\n return ans;\n}\n\n};\n" }, { "language": "py", - "text": "\nclass Solution:\n def trap(self, heights: List[int]) -> int:\n n = len(heights)\n l, r = [0] * (n + 1), [0] * (n + 1)\n ans = 0\n for i in range(1, len(heights) + 1):\n l[i] = max(l[i - 1], heights[i - 1])\n for i in range(len(heights) - 1, 0, -1):\n r[i] = max(r[i + 1], heights[i])\n for i in range(len(heights)):\n ans += max(0, min(l[i + 1], r[i]) - heights[i])\n return ans \n" + "text": "\nclass Solution:\n def trap(self, heights: List[int]) -> int:\n n = len(heights)\n l, r = [0] * (n + 1), [0] * (n + 1)\n ans = 0\n for i in range(1, len(heights) + 1):\n l[i] = max(l[i - 1], heights[i - 1])\n for i in range(len(heights) - 1, 0, -1):\n r[i] = max(r[i + 1], heights[i])\n for i in range(len(heights)):\n ans += max(0, min(l[i + 1], r[i]) - heights[i])\n return ans\n" + }, + { + "language": "py", + "text": "\nclass Solution:\n def trap(self, heights: List[int]) -> int:\n n = len(heights)\n l_max = r_max = 0\n l, r = 0, n - 1\n ans = 0\n while l < r:\n if heights[l] < heights[r]:\n if heights[l] < l_max:\n ans += l_max - heights[l]\n else:\n l_max = heights[l]\n l += 1\n else:\n if heights[r] < r_max:\n ans += r_max - heights[r]\n else:\n r_max = heights[r]\n r -= 1\n return ans\n" } ] }, @@ -5999,6 +6062,87 @@ } ] }, +"unique-paths-ii":{ + "id": "63", + "name": "unique-paths-ii", + "company": [], + "pre": [ + { + "text": " 动态规划", + "link": null, + "color": "red" + } + ], + "keyPoints": [ + { + "text": "记忆化递归", + "link": null, + "color": "blue" + }, + { + "text": "基本动态规划问题", + "link": null, + "color": "blue" + }, + { + "text": "空间复杂度可以进一步优化到O(n),这会是一个考点", + "link": null, + "color": "blue" + } + ], + "solution": "https://github.com/azl397985856/leetcode/blob/master/problems/63.unique-paths-ii.md", + "code": [ + { + "language": "py", + "text": "\nclass Solution:\n def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:\n m = len(obstacleGrid)\n n = len(obstacleGrid[0])\n if obstacleGrid[0][0]:\n return 0\n\n dp = [[0] * (n + 1) for _ in range(m + 1)]\n dp[1][1] = 1\n\n for i in range(1, m + 1):\n for j in range(1, n + 1):\n if i == 1 and j == 1:\n continue\n if obstacleGrid[i - 1][j - 1] == 0:\n dp[i][j] = dp[i - 1][j] + dp[i][j - 1]\n else:\n dp[i][j] = 0\n return dp[m][n]\n" + }, + { + "language": "py", + "text": "\nclass Solution:\n def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:\n m = len(obstacleGrid)\n n = len(obstacleGrid[0])\n if obstacleGrid[0][0]:\n return 0\n\n dp = [0] * (n + 1)\n dp[1] = 1\n for i in range(1, m + 1):\n for j in range(1, n + 1):\n if obstacleGrid[i - 1][j - 1] == 0:\n dp[j] += dp[j - 1]\n else:\n dp[j] = 0\n return dp[-1]\n" + } + ] +}, +"maximum-length-of-repeated-subarray":{ + "id": "718", + "name": "maximum-length-of-repeated-subarray", + "company": [], + "pre": [ + { + "text": " 哈希表", + "link": null, + "color": "red" + }, + { + "text": " 数组", + "link": null, + "color": "red" + }, + { + "text": " 二分查找", + "link": null, + "color": "red" + }, + { + "text": " 动态规划", + "link": null, + "color": "red" + } + ], + "keyPoints": [ + { + "text": "dp建模套路", + "link": null, + "color": "blue" + } + ], + "solution": "https://github.com/azl397985856/leetcode/blob/master/problems/718.maximum-length-of-repeated-subarray.md", + "code": [ + { + "language": "py", + "text": "\nclass Solution:\n def findLength(self, A, B):\n m, n = len(A), len(B)\n ans = 0\n dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]\n for i in range(1, m + 1):\n for j in range(1, n + 1):\n if A[i - 1] == B[j - 1]:\n dp[i][j] = dp[i - 1][j - 1] + 1\n ans = max(ans, dp[i][j])\n return ans\n" + } + ] +}, "accounts-merge":{ "id": "721", "name": "accounts-merge", @@ -6148,6 +6292,42 @@ } ] }, +"is-graph-bipartite":{ + "id": "785", + "name": "is-graph-bipartite", + "company": [], + "pre": [ + { + "text": " 图的遍历", + "link": null, + "color": "red" + }, + { + "text": " DFS", + "link": null, + "color": "red" + } + ], + "keyPoints": [ + { + "text": "图的建立和遍历", + "link": null, + "color": "blue" + }, + { + "text": "colors数组", + "link": null, + "color": "blue" + } + ], + "solution": "https://github.com/azl397985856/leetcode/blob/master/problems/785.is-graph-bipartite.md", + "code": [ + { + "language": "py", + "text": "\nclass Solution:\n def dfs(self, grid, colors, i, color, N):\n colors[i] = color\n for j in range(N):\n if grid[i][j] == 1:\n if colors[j] == color:\n return False\n if colors[j] == 0 and not self.dfs(grid, colors, j, -1 * color, N):\n return False\n return True\n\n def isBipartite(self, graph: List[List[int]]) -> bool:\n N = len(graph)\n grid = [[0] * N for _ in range(N)]\n colors = [0] * N\n for i in range(N):\n for j in graph[i]:\n grid[i][j] = 1\n for i in range(N):\n if colors[i] == 0 and not self.dfs(grid, colors, i, 1, N):\n return False\n return True\n" + } + ] +}, "word-search-en":{ "id": "79", "name": "word-search-en", @@ -6158,7 +6338,7 @@ "code": [ { "language": "java", - "text": "\npublic class LC79WordSearch {\n public boolean exist(char[][] board, String word) {\n if (board == null || board.length == 0 || board[0].length == 0\n || word == null || word.length() == 0) return true;\n int rows = board.length;\n int cols = board[0].length;\n for (int r = 0; r < rows; r++) {\n for (int c = 0; c < cols; c++) {\n // scan board, start with word first character \n if (board[r][c] == word.charAt(0)) {\n if (helper(board, word, r, c, 0)) {\n return true;\n }\n }\n }\n }\n return false;\n }\n \n private boolean helper(char[][] board, String word, int r, int c, int start) {\n // already match word all characters, return true\n if (start == word.length()) return true;\n if (!isValid(board, r, c) ||\n board[r][c] != word.charAt(start)) return false;\n // mark visited\n board[r][c] = '*';\n boolean res = helper(board, word, r + 1, c, start + 1)\n || helper(board, word, r, c + 1, start + 1)\n || helper(board, word, r - 1, c, start + 1)\n || helper(board, word, r, c - 1, start + 1);\n // backtracking to start position\n board[r][c] = word.charAt(start);\n return res;\n }\n \n private boolean isValid(char[][] board, int r, int c) {\n return r >= 0 && r < board.length && c >= 0 && c < board[0].length;\n }\n}\n" + "text": "\npublic class LC79WordSearch {\n public boolean exist(char[][] board, String word) {\n if (board == null || word == null) return false;\n if (word.length() == 0) return true;\n if (board.length == 0) return false;\n int rows = board.length;\n int cols = board[0].length;\n for (int r = 0; r < rows; r++) {\n for (int c = 0; c < cols; c++) {\n // scan board, start with word first character \n if (board[r][c] == word.charAt(0)) {\n if (helper(board, word, r, c, 0)) {\n return true;\n }\n }\n }\n }\n return false;\n }\n \n private boolean helper(char[][] board, String word, int r, int c, int start) {\n // already match word all characters, return true\n if (start == word.length()) return true;\n if (!isValid(board, r, c) ||\n board[r][c] != word.charAt(start)) return false;\n // mark visited\n board[r][c] = '*';\n boolean res = helper(board, word, r + 1, c, start + 1)\n || helper(board, word, r, c + 1, start + 1)\n || helper(board, word, r - 1, c, start + 1)\n || helper(board, word, r, c - 1, start + 1);\n // backtracking to start position\n board[r][c] = word.charAt(start);\n return res;\n }\n \n private boolean isValid(char[][] board, int r, int c) {\n return r >= 0 && r < board.length && c >= 0 && c < board[0].length;\n }\n}\n" }, { "language": "js", @@ -6207,7 +6387,7 @@ "code": [ { "language": "java", - "text": "\npublic class LC79WordSearch {\n public boolean exist(char[][] board, String word) {\n if (board == null || board.length == 0 || board[0].length == 0\n || word == null || word.length() == 0) return true;\n int rows = board.length;\n int cols = board[0].length;\n for (int r = 0; r < rows; r++) {\n for (int c = 0; c < cols; c++) {\n // scan board, start with word first character\n if (board[r][c] == word.charAt(0)) {\n if (helper(board, word, r, c, 0)) {\n return true;\n }\n }\n }\n }\n return false;\n }\n\n private boolean helper(char[][] board, String word, int r, int c, int start) {\n // already match word all characters, return true\n if (start == word.length()) return true;\n if (!isValid(board, r, c) ||\n board[r][c] != word.charAt(start)) return false;\n // mark visited\n board[r][c] = '*';\n boolean res = helper(board, word, r - 1, c, start + 1) // 上\n || helper(board, word, r + 1, c, start + 1) // 下\n || helper(board, word, r, c - 1, start + 1) // 左\n || helper(board, word, r, c + 1, start + 1); // 右\n // backtracking to start position\n board[r][c] = word.charAt(start);\n return res;\n }\n\n private boolean isValid(char[][] board, int r, int c) {\n return r >= 0 && r < board.length && c >= 0 && c < board[0].length;\n }\n}\n" + "text": "\npublic class LC79WordSearch {\n public boolean exist(char[][] board, String word) {\n if (board == null || word == null) return false;\n if (word.length() == 0) return true;\n if (board.length == 0) return false;\n int rows = board.length;\n int cols = board[0].length;\n for (int r = 0; r < rows; r++) {\n for (int c = 0; c < cols; c++) {\n // scan board, start with word first character\n if (board[r][c] == word.charAt(0)) {\n if (helper(board, word, r, c, 0)) {\n return true;\n }\n }\n }\n }\n return false;\n }\n\n private boolean helper(char[][] board, String word, int r, int c, int start) {\n // already match word all characters, return true\n if (start == word.length()) return true;\n if (!isValid(board, r, c) ||\n board[r][c] != word.charAt(start)) return false;\n // mark visited\n board[r][c] = '*';\n boolean res = helper(board, word, r - 1, c, start + 1) // 上\n || helper(board, word, r + 1, c, start + 1) // 下\n || helper(board, word, r, c - 1, start + 1) // 左\n || helper(board, word, r, c + 1, start + 1); // 右\n // backtracking to start position\n board[r][c] = word.charAt(start);\n return res;\n }\n\n private boolean isValid(char[][] board, int r, int c) {\n return r >= 0 && r < board.length && c >= 0 && c < board[0].length;\n }\n}\n" }, { "language": "js", @@ -6443,9 +6623,25 @@ ], "solution": "https://github.com/azl397985856/leetcode/blob/master/problems/875.koko-eating-bananas.md", "code": [ + { + "language": "java", + "text": "\npublic int binarySearch(int[] nums, int target) {\n // 左右都闭合的区间 [l, r]\n int left = 0; \n int right = nums.length - 1;\n\n while(left <= right) {\n int mid = left + (right - left) / 2;\n if(nums[mid] == target)\n return mid; \n else if (nums[mid] < target)\n\t\t\t// 搜索区间变为 [mid+1, right]\n left = mid + 1;\n else if (nums[mid] > target)\n // 搜索区间变为 [left, mid - 1]\n right = mid - 1;\n }\n return -1;\n}\n" + }, + { + "language": "java", + "text": "\npublic int binarySearchLeft(int[] nums, int target) {\n\t// 搜索区间为 [left, right]\n int left = 0;\n int right = nums.length - 1;\n while (left <= right) {\n int mid = left + (right - left) / 2;\n if (nums[mid] < target) {\n // 搜索区间变为 [mid+1, right]\n left = mid + 1;\n } else if (nums[mid] > target) {\n // 搜索区间变为 [left, mid-1]\n right = mid - 1;\n } else if (nums[mid] == target) {\n // 收缩右边界\n right = mid - 1;\n }\n }\n // 检查是否越界\n if (left >= nums.length || nums[left] != target)\n return -1;\n return left;\n}\n" + }, + { + "language": "java", + "text": "\npublic int binarySearchRight(int[] nums, int target) {\n\t// 搜索区间为 [left, right]\n int left = 0\n int right = nums.length - 1;\n while (left <= right) {\n int mid = left + (right - left) / 2;\n if (nums[mid] < target) {\n\t\t\t// 搜索区间变为 [mid+1, right]\n left = mid + 1;\n } else if (nums[mid] > target) {\n\t\t\t// 搜索区间变为 [left, mid-1]\n right = mid - 1;\n } else if (nums[mid] == target) {\n // 收缩左边界\n left = mid + 1;\n }\n }\n // 检查是否越界\n if (right < 0 || nums[right] != target)\n return -1;\n return right;\n}\n" + }, { "language": "js", - "text": "\n\nfunction canEatAllBananas(piles, H, mid) {\n let h = 0;\n for(let pile of piles) {\n h += Math.ceil(pile / mid);\n }\n\n return h <= H;\n }\n/**\n * @param {number[]} piles\n * @param {number} H\n * @return {number}\n */\nvar minEatingSpeed = function(piles, H) {\n let lo = 1,\n hi = Math.max(...piles);\n\n while(lo <= hi) {\n let mid = lo + ((hi - lo) >> 1);\n if (canEatAllBananas(piles, H, mid)) {\n hi = mid - 1;\n } else {\n lo = mid + 1;\n }\n }\n\n return lo; // 不能选择hi\n};\n" + "text": "\n\nfunction canEatAllBananas(piles, H, mid) {\n let h = 0;\n for(let pile of piles) {\n h += Math.ceil(pile / mid);\n }\n\n return h <= H;\n }\n/**\n * @param {number[]} piles\n * @param {number} H\n * @return {number}\n */\nvar minEatingSpeed = function(piles, H) {\n let lo = 1,\n hi = Math.max(...piles);\n // [l, r) ， 左闭右开的好处是如果能找到，那么返回 l 和 r 都是一样的，因为最终 l 等于 r。\n while(lo <= hi) {\n let mid = lo + ((hi - lo) >> 1);\n if (canEatAllBananas(piles, H, mid)) {\n hi = mid - 1;\n } else {\n lo = mid + 1;\n }\n }\n\n return lo; // 不能选择hi\n};\n" + }, + { + "language": "py", + "text": "\nclass Solution:\n def canEatAllBananas(self, piles, H, K):\n t = 0\n for pile in piles:\n t += math.ceil(pile / K)\n return t <= H\n def minEatingSpeed(self, piles: List[int], H: int) -> int:\n l, r = 1, max(piles)\n # [l, r) ， 左闭右开的好处是如果能找到，那么返回 l 和 r 都是一样的，因为最终 l 等于 r。 \n while l < r:\n mid = (l + r) >> 1\n if self.canEatAllBananas(piles, H, mid):\n r = mid\n else:\n l = mid + 1\n return l\n\n" } ] }, @@ -6518,6 +6714,60 @@ } ] }, +"possible-bipartition":{ + "id": "886", + "name": "possible-bipartition", + "company": [], + "pre": [ + { + "text": " 图的遍历", + "link": null, + "color": "red" + }, + { + "text": " DFS", + "link": null, + "color": "red" + } + ], + "keyPoints": [ + { + "text": "二分图", + "link": null, + "color": "blue" + }, + { + "text": "染色法", + "link": null, + "color": "blue" + }, + { + "text": "图的建立和遍历", + "link": null, + "color": "blue" + }, + { + "text": "colors数组", + "link": null, + "color": "blue" + } + ], + "solution": "https://github.com/azl397985856/leetcode/blob/master/problems/886.possible-bipartition.md", + "code": [ + { + "language": "py", + "text": "\n graph = [[0] * N for i in range(N)]\n for a, b in dislikes:\n graph[a - 1][b - 1] = 1\n graph[b - 1][a - 1] = 1\n" + }, + { + "language": "py", + "text": "\n# 其中j 表示当前是第几个人，N表示总人数。 dfs的功能就是根据colors和graph分配组，true表示可以分，false表示不可以，具体代码见代码区。\nif colors[j] == 0 and not self.dfs(graph, colors, j, -1 * color, N)\n" + }, + { + "language": "py", + "text": "\nclass Solution:\n def dfs(self, graph, colors, i, color, N):\n colors[i] = color\n for j in range(N):\n # dislike eachother\n if graph[i][j] == 1:\n if colors[j] == color:\n return False\n if colors[j] == 0 and not self.dfs(graph, colors, j, -1 * color, N):\n return False\n return True\n\n def possibleBipartition(self, N: int, dislikes: List[List[int]]) -> bool:\n graph = [[0] * N for i in range(N)]\n colors = [0] * N\n for a, b in dislikes:\n graph[a - 1][b - 1] = 1\n graph[b - 1][a - 1] = 1\n for i in range(N):\n if colors[i] == 0 and not self.dfs(graph, colors, i, 1, N):\n return False\n return True\n\n" + } + ] +}, "super-egg-drop":{ "id": "887", "name": "super-egg-drop", @@ -6731,6 +6981,11 @@ } ], "keyPoints": [ + { + "text": "四点法", + "link": null, + "color": "blue" + }, { "text": "链表的基本操作", "link": null, @@ -6758,17 +7013,13 @@ "language": "js", "text": "\n/*\n * @lc app=leetcode id=92 lang=javascript\n *\n * [92] Reverse Linked List II\n *\n * https://leetcode.com/problems/reverse-linked-list-ii/description/\n */\n/**\n * Definition for singly-linked list.\n * function ListNode(val) {\n * this.val = val;\n * this.next = null;\n * }\n */\n/**\n * @param {ListNode} head\n * @param {number} m\n * @param {number} n\n * @return {ListNode}\n */\nvar reverseBetween = function (head, m, n) {\n // 虚拟节点，简化操作\n const dummyHead = {\n next: head,\n };\n\n let cur = dummyHead.next; // 当前遍历的节点\n let pre = cur; // 因为要反转，因此我们需要记住前一个节点\n let index = 0; // 链表索引，用来判断是否是特殊位置（头尾位置）\n\n // 上面提到的四个特殊节点\n let p1 = (p2 = p3 = p4 = null);\n\n while (cur) {\n const next = cur.next;\n index++;\n\n // 对 (m - n) 范围内的节点进行反转\n if (index > m && index <= n) {\n cur.next = pre;\n }\n\n // 下面四个if都是边界, 用于更新四个特殊节点的值\n if (index === m - 1) {\n p1 = cur;\n }\n if (index === m) {\n p2 = cur;\n }\n\n if (index === n) {\n p3 = cur;\n }\n\n if (index === n + 1) {\n p4 = cur;\n }\n\n pre = cur;\n\n cur = next;\n }\n\n // 两个链表合并起来\n (p1 || dummyHead).next = p3; // 特殊情况需要考虑\n p2.next = p4;\n\n return dummyHead.next;\n};\n" }, - { - "language": "cpp", - "text": "\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n * int val;\n * ListNode *next;\n * ListNode(int x) : val(x), next(NULL) {}\n * };\n */\nclass Solution {\npublic:\n ListNode* reverseBetween(ListNode* head, int s, int e) {\n if (s == e) return head;\n ListNode* prev = nullptr;\n auto cur = head;\n for (int i = 1; i < s; ++i) {\n prev = cur;\n cur = cur->next;\n }\n // 此时各指针指向：\n // x -> x -> x -> x -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> x -> x -> x ->\n // ^head ^prev ^cur\n ListNode* p = nullptr;\n auto c = cur;\n auto tail = c;\n ListNode* n = nullptr;\n for (int i = s; i <= e; ++i) {\n n = c->next;\n c->next = p;\n p = c;\n c = n;\n }\n // 此时各指针指向：\n // x -> x -> x -> x 8 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1 x -> x -> x ->\n // ^head ^prev ^p ^cur ^c\n // ^tail\n if (prev != nullptr) { // 若指向前一个节点的指针不为空，则说明s在链表中间（不是头节点）\n prev->next = p;\n cur->next = c;\n return head;\n } else {\n if (tail != nullptr) tail->next = c;\n return p;\n }\n }\n};\n" - }, { "language": "py", - "text": "\nclass Solution:\n def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:\n pre = None\n cur = head\n i = 0\n p1 = p2 = p3 = p4 = None\n # ...\n if p1:\n p1.next = p3\n else:\n dummy.next = p3\n if p2:\n p2.next = p4\n return head\n" + "text": "\nclass Solution:\n def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:\n pre = None\n cur = head\n i = 0\n p1 = p2 = p3 = p4 = None\n # 一坨逻辑\n if p1:\n p1.next = p3\n else:\n dummy.next = p3\n if p2:\n p2.next = p4\n return head\n" }, { "language": "py", - "text": "\nclass Solution:\n def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:\n pre = None\n cur = head\n i = 0\n p1 = p2 = p3 = p4 = None\n dummy = ListNode(0)\n dummy.next = head\n\n # ...\n\n if p1:\n p1.next = p3\n else:\n dummy.next = p3\n if p2:\n p2.next = p4\n\n return dummy.next\n" + "text": "\nclass Solution:\n def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:\n pre = None\n cur = head\n i = 0\n p1 = p2 = p3 = p4 = None\n dummy = ListNode(0)\n dummy.next = head\n # 一坨逻辑\n if p1:\n p1.next = p3\n else:\n dummy.next = p3\n if p2:\n p2.next = p4\n\n return dummy.next\n" }, { "language": "py", @@ -6855,7 +7106,7 @@ }, { "language": "js", - "text": "\n/*\n * @lc app=leetcode id=94 lang=javascript\n *\n * [94] Binary Tree Inorder Traversal\n *\n * https://leetcode.com/problems/binary-tree-inorder-traversal/description/\n *\n * algorithms\n * Medium (55.22%)\n * Total Accepted: 422.4K\n * Total Submissions: 762.1K\n * Testcase Example: '[1,null,2,3]'\n *\n * Given a binary tree, return the inorder traversal of its nodes' values.\n *\n * Example:\n *\n *\n * Input: [1,null,2,3]\n * ⁠ 1\n * ⁠ \\\n * ⁠ 2\n * ⁠ /\n * ⁠ 3\n *\n * Output: [1,3,2]\n *\n * Follow up: Recursive solution is trivial, could you do it iteratively?\n *\n */\n/**\n * Definition for a binary tree node.\n * function TreeNode(val) {\n * this.val = val;\n * this.left = this.right = null;\n * }\n */\n/**\n * @param {TreeNode} root\n * @return {number[]}\n */\nvar inorderTraversal = function (root) {\n // 1. Recursive solution\n // if (!root) return [];\n // const left = root.left ? inorderTraversal(root.left) : [];\n // const right = root.right ? inorderTraversal(root.right) : [];\n // return left.concat([root.val]).concat(right);\n\n // 2. iterative solutuon\n if (!root) return [];\n const stack = [root];\n const ret = [];\n let left = root.left;\n\n let item = null; // stack 中弹出的当前项\n\n while (left) {\n stack.push(left);\n left = left.left;\n }\n\n while ((item = stack.pop())) {\n ret.push(item.val);\n let t = item.right;\n\n while (t) {\n stack.push(t);\n t = t.left;\n }\n }\n\n return ret;\n};\n" + "text": "\n/**\n * @param {TreeNode} root\n * @return {number[]}\n */\nvar inorderTraversal = function (root) {\n // 1. Recursive solution\n // if (!root) return [];\n // const left = root.left ? inorderTraversal(root.left) : [];\n // const right = root.right ? inorderTraversal(root.right) : [];\n // return left.concat([root.val]).concat(right);\n\n // 2. iterative solutuon\n if (!root) return [];\n const stack = [root];\n const ret = [];\n let left = root.left;\n\n let item = null; // stack 中弹出的当前项\n\n while (left) {\n stack.push(left);\n left = left.left;\n }\n\n while ((item = stack.pop())) {\n ret.push(item.val);\n let t = item.right;\n\n while (t) {\n stack.push(t);\n t = t.left;\n }\n }\n\n return ret;\n};\n" }, { "language": "cpp",