weixin_38199770的博客

官方题解//解法一：判断左右两边哪个长度为奇数，答案就在哪边class Solution { public int singleNonDuplicate(int[] nums) { int l = 0; int r = nums.length - 1; int L = 0;//左边长度 int R = 0;//右边长度 while(l < r){ int mid = l + r &gt.

//解法一：自己写出来的基本的动态规划class Solution { public int lengthOfLIS(int[] nums) { int n = nums.length; int[] dp = new int[n]; dp[0] = 1; int ans = 1; for(int i = 1; i < n; i++){ int len = 0; f.

//解法一：排序+二分 自己写的 好慢。。。。class Solution { public int findDuplicate(int[] nums) { Arrays.sort(nums); for(int i = 0; i < nums.length; i++) System.out.print(nums[i] + " "); int n = nums.length; int l = 0; .

https://blog.youkuaiyun.com/u011646912/article/details/102580812（转载）1.题目描述给定一个不为空的二叉搜索树和一个目标值 target，请在该二叉搜索树中找到最接近目标值 target 的数值。注意：给定的目标值 target 是一个浮点数题目保证在该二叉搜索树中只会存在一个最接近目标值的数示例：输入: root = [4,2,5,1,3]，目标值 target = 3.714286 4 / \ 2 5...

//其实能做出来的，起始的边界没想清楚public class Solution { // 以下的代码注释是 @coder_hezi 帮助添加的，在此表示感谢 public int hIndex(int[] citations) { int len = citations.length; if (len == 0 || citations[len - 1] == 0) { return 0; } .

//写法一：逐行二分 可提前判断是否有解class Solution { public boolean searchMatrix(int[][] matrix, int target) { for(int i = 0; i < matrix.length; i++){ if(matrix[i][0] > target) break; if(matrix[i][matrix[i].len..

官方题解//解法一:利用中序遍历将值存储在数组里class Solution { public ArrayList<Integer> inorder(TreeNode root, ArrayList<Integer> arr) { if (root == null) return arr; inorder(root.left, arr); arr.add(root.val); inorder(root.right, arr); ret.

官方二分解法//二分解法 真没想到class Solution { public int countNodes(TreeNode root) { if (root == null) { return 0; } int level = 0; TreeNode node = root; while (node.left != null) { level++; .

//解法一：前缀和 + 二分//https://leetcode-cn.com/problems/minimum-size-subarray-sum/solution/qian-zhui-he-er-fen-hua-dong-chuang-kou-by-coldme-/class Solution { public int minSubArrayLen(int s, int[] nums) { int n = nums.length; int[] sum = n.

//解法一：二分+正向dp//https://leetcode-cn.com/problems/dungeon-game/solution/er-fen-cwo-te-yao-sou-bao-by-tian-xin-yue-yuan-3/class Solution { private boolean search(int[][] d, int hp){ int m = d.length; int n = d[0].length; int[][.

大佬题解.//解法一:分治class Solution { public int findMin(int[] nums) { int n = nums.length; int l = 0; int r = n - 1; return bsearch(nums, l, r); }/*发现了一个规律 递增的旋转数组找最小值 mid 只能与 right比较*/ private int bsearch(int[] .

【大佬题解】https://leetcode-cn.com/problems/find-minimum-in-rotated-sorted-array/solution/er-fen-fa-fen-zhi-fa-python-dai-ma-java-dai-ma-by-///解法一：mid只能与right比较不能与left比较public class Solution { public int findMin(int[] nums) { int len = nums.leng.

【大佬题解】https://leetcode-cn.com/problems/search-in-rotated-sorted-array-ii/solution/er-fen-cha-zhao-by-liweiwei1419///解法一：mid 与 left 比较分类讨论public class Solution { public boolean search(int[] nums, int target) { int len = nums.length; .

//解法一: 一次二分 需要考虑好多边界class Solution { public int search(int[] nums, int target) { int n = nums.length; int l = 0; int r = n - 1; while(l < r){ int mid = l + r >> 1; if(nums[mid] >= nu.

//快速乘法+二分//另附快速乘法和快速幂模板class Solution { public int divide(int a, int b) { long x = a, y = b; boolean isNeg = false; if ((x > 0 && y < 0) || (x < 0 && y > 0)) isNeg = true; if (x < 0) x =.

//解法一://转化为找第k小的数class Solution { public double findMedianSortedArrays(int[] nums1, int[] nums2) { int n = nums1.length; int m = nums2.length; int tol = n + m; if(tol % 2 == 1)//奇数找第 tol / 2 + 1的数 return d.