commit

array/11_盛最多水的容器/11_medium_盛最多水的容器.py

@@ -36,21 +36,27 @@
 # @lc code=start
 class Solution:
     def maxArea(self, height: List[int]) -> int:
-        Area = 0
-        i, j = 0, len(height) - 1
-        while i < j:
-            if (height[i] < height[j]) :
-                minHeight = height[i]
-                i += 1
-                area = (j-i+1) * minHeight
-                Area = max(area, Area)
+        """求取最大面积区域
+        Args:
+            height: a list with height
+        Returns:
+            The maxArea
+        """
+        # 双指针解法
+        max_area = 0
+        leftPointer, rightPointer = 0, len(height) - 1
+        while leftPointer < rightPointer:
+            if height[leftPointer] < height[rightPointer]:
+                area = height[leftPointer] * (rightPointer - leftPointer)
+                leftPointer += 1
+                max_area = max(area, max_area)
             else:
-                minHeight = height[j]
-                j -= 1
-                area = (j-i+1) * minHeight
-                Area = max(area, Area)
-        return Area
-
+                area = height[rightPointer] * (rightPointer - leftPointer)
+                rightPointer -= 1
+                max_area = max(area, max_area)
+        return max_area
+
+       
 
 # @lc code=end
 

array/4_寻找两个有序数组的中位数/4_hard_寻找两个有序数组的中位数.py

@@ -0,0 +1,46 @@
+#
+# @lc app=leetcode.cn id=4 lang=python3
+#
+# [4] 寻找两个有序数组的中位数
+#
+# https://leetcode-cn.com/problems/median-of-two-sorted-arrays/description/
+#
+# algorithms
+# Hard (37.27%)
+# Likes:    2822
+# Dislikes: 0
+# Total Accepted:    215.9K
+# Total Submissions: 563.6K
+# Testcase Example:  '[1,3]\n[2]'
+#
+# 给定两个大小为 m 和 n 的正序（从小到大）数组 nums1 和 nums2。
+# 
+# 请你找出这两个正序数组的中位数，并且要求算法的时间复杂度为 O(log(m + n))。
+# 
+# 你可以假设 nums1 和 nums2 不会同时为空。
+# 
+# 
+# 
+# 示例 1:
+# 
+# nums1 = [1, 3]
+# nums2 = [2]
+# 
+# 则中位数是 2.0
+# 
+# 
+# 示例 2:
+# 
+# nums1 = [1, 2]
+# nums2 = [3, 4]
+# 
+# 则中位数是 (2 + 3)/2 = 2.5
+# 
+# 
+#
+
+# @lc code=start
+class Solution:
+    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
+# @lc code=end
+

hash-table/1_两数之和/1_easy_两数之和.py

@@ -38,7 +38,6 @@ def twoSum(self, nums, target) :
                 return [temp_dict[nums[i]], i]
             else:   
                 temp_dict[target - nums[i]] = i
-
 Solution().twoSum([2, 7, 11, 15], 13)
 
 # @lc code=end

linked-list/25_K 个一组翻转链表/25_hard_k个一组翻转链表.py

@@ -48,39 +48,35 @@
 #         self.next = None
 
 class Solution:
-    # 翻转一个子链表，并且返回新的头与尾
-    def reverse(self, head: ListNode, tail: ListNode):
-        prev = tail.next
-        p = head
-        while prev != tail:
-            nex = p.next
-            p.next = prev
-            prev = p
-            p = nex
-        return tail, head
-
     def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
-        hair = ListNode(0)
-        hair.next = head
-        pre = hair
+        dummy = ListNode(-1)
+        dummy.next = head
+        start = dummy
 
-        while head:
-            tail = pre
-            # 查看剩余部分长度是否大于等于 k
+        while start:
+            # 设置end结点，看是否够k个一组
+            end = start
             for i in range(k):
-                tail = tail.next
-                if not tail:
-                    return hair.next
-            nex = tail.next
-            head, tail = self.reverse(head, tail)
-            # 把子链表重新接回原链表
-            pre.next = head
-            tail.next = nex
-            pre = tail
-            head = tail.next
+                end = end.next
+                if not end:
+                    return dummy.next
+
+            # swap node
+            swap_a = start.next
+            swap_b = swap_a.next
+            for i in range(k - 1):
+                temp = swap_b.next
+                swap_b.next = swap_a
+                swap_a = swap_b
+                swap_b = temp
+            temp = start.next
+            start.next = swap_a
+            temp.next = swap_b
+            start = temp
+        return dummy.next
 
-        return hair.next
 
 
+
 # @lc code=end