fd

Author: dingjikerbo

README.md

@@ -387,6 +387,7 @@
 |654|[Maximum Binary Tree](https://leetcode.com/problems/maximum-binary-tree/description/)|[Java](leetcode/solution/src/MaximumBinaryTree.java)|100||
 |669|[Trim a Binary Search Tree](https://leetcode.com/problems/trim-a-binary-search-tree/)|[Java](leetcode/solution/src/TrimABinarySearchTree.java)|100||
 |681|[Next Closest Time](https://leetcode.com/problems/next-closest-time/)|[Java](leetcode/solution/src/NextClosestTime.java)|80|这题多做几遍|
+|683|[K Empty Slots](https://leetcode.com/problems/k-empty-slots/)|[Java](leetcode/solution/src/KEmptySlots.java)|80||
 |684|[Redundant Connection](https://leetcode.com/problems/redundant-connection/)| [Java](leetcode/solution/src/RedundantConnection.java)|||
 |695|[Max Area of Island](https://leetcode.com/problems/max-area-of-island/)|[Java](leetcode/solution/src/MaxAreaOfIsland.java)|100||
 |703|[Kth Largest Element in a Stream](https://leetcode.com/problems/kth-largest-element-in-a-stream/)|[Java](leetcode/solution/src/KthLargestElementInAStream.java)|100||

leetcode/solution/src/KEmptySlots.java

@@ -0,0 +1,58 @@
+import java.util.TreeSet;
+
+/**
+ * 这题意思是第i盆花第flowers[i]天开，问第几天的时候存在这么一种情况：有两盆已经开的花中间恰好隔着k盆没开的花
+ */
+public class KEmptySlots {
+
+    public int kEmptySlots(int[] flowers, int k) {
+        TreeSet<Integer> map = new TreeSet<>();
+        for (int i = 0; i < flowers.length; i++) {
+            int flower = flowers[i] - 1;
+            map.add(flower);
+
+            Integer key = map.ceiling(flower + 1);
+            if (key != null && key - flower == k + 1) {
+                return i + 1;
+            }
+            key = map.floor(flower - 1);
+            if (key != null && flower - key == k + 1) {
+                return i + 1;
+            }
+        }
+
+        return -1;
+    }
+
+    /**
+     * 时间复杂度O(n)
+     * days表示第i盆花第几天开
+     * 天数越大，表示花开的越晚
+     * 对于(left, right)这个区间，(这个区间刚好相隔k)，如果days值都比left和right大，说明(left,right)区间的花都开的比left和right晚
+     * 换言之，在left和right开的时候，中间的花都没开，而left和right刚好相隔k，符合条件。
+     * 我们选择left和right之中较大的那个值作为备选结果之一，选取较大是因为表示发生的较晚，刚好触发条件
+     * 但是题目要返回最早的那一天，所以我们要继续遍历下去
+     *
+     * 此外还要注意，当发现(left,right)区间存在不符合条件的情况，则sliding window要从i开始，因为如果从(left,i)区间的任何一个数开始，都不会符合条件
+     * 另外当找到一组符合条件的情况后，sliding window也要从i开始，也就是从right开始，假如从left+1开始，则当前的right就不符合大于sliding window两端的条件了
+     */
+    public int kEmptySlots2(int[] flowers, int k) {
+        int[] days = new int[flowers.length];
+        for (int i = 0; i < flowers.length; i++) days[flowers[i] - 1] = i + 1;
+        int left = 0, right = k + 1, res = Integer.MAX_VALUE;
+        for (int i = left + 1; i <= right && right < days.length; i++) {
+            if (i == right) {
+                res = Math.min(res, Math.max(days[left], days[right]));
+                left = i;
+                right = k + 1 + i;
+                continue;
+            }
+
+            if (days[i] < days[left] || days[i] < days[right]) {
+                left = i;
+                right = k + 1 + i;
+            }
+        }
+        return (res == Integer.MAX_VALUE) ? -1 : res;
+    }
+}

leetcode/src/Main.java

@@ -1,36 +1,46 @@
+import com.sun.org.apache.xpath.internal.operations.Bool;
+
 import java.util.*;
 
 public class Main {
 
     public static class Solution {
 
-        public String nearestPalindromic(String n) {
-            long val = Long.parseLong(n);
-            for (int i = 0; ; i++) {
-                long k1 = val - i, k2 = val + i;
-                if (isPalindrome(k1)) {
-                    return String.valueOf(k1);
-                }
-                if (isPalindrome(k2)) {
-                    return String.valueOf(k2);
+        public int kEmptySlots(int[] flowers, int k) {
+            int[] days = new int[flowers.length];
+            for (int i = 0; i < flowers.length; i++) days[flowers[i] - 1] = i + 1;
+            int left = 0, right = k + 1, res = Integer.MAX_VALUE;
+            for (int i = left + 1; i <= right && right < days.length; i++) {
+                if (i == right) {
+                    res = Math.min(res, Math.max(days[left], days[right]));
                 }
-            }
-        }
 
-        private boolean isPalindrome(long k) {
-            long x = k, rev = 0;
-            for ( ; k > 0; ) {
-                rev = rev * 10 + k % 10;
-                k /= 10;
+                if (days[i] < days[left] || days[i] < days[right]) {
+                    left = i;
+                    right = k + 1 + i;
+                }
             }
-            return rev == x;
+            return (res == Integer.MAX_VALUE) ? -1 : res;
         }
 
     }
 
     public static void main(String[] args) {
         Solution solution = new Solution();
-        String s = solution.nearestPalindromic("807045053224792883");
-        System.out.println(s);
+
+        TreeMap<Integer, Boolean> map = new TreeMap<>();
+        int[] nums = {4, 1, 7, 5, 2, 8, 10, 0};
+        for (int n : nums) {
+            map.put(n, true);
+        }
+        for (int k : map.keySet()) {
+            System.out.print(k + " ");
+        }
+
+        int n = solution.kEmptySlots(new int[]{
+                1, 2, 3
+        }, 1);
+        System.out.println(n);
+
     }
 }