for (int i = 0; i < inorder.size(); i++) { hash[inorder[i]] = i; } return helper(0, 0, inorder.size() - 1); }private: vector<int> preorder; unordered_map<int, int> hash; TreeNode* helper(int root, int left, int right) {if...
public class Solution { public boolean IsPopOrder(int [] pushA,int [] popA) { Stack<Integer> stack = new Stack<>(); int l = pushA.length; int j = 0; for(int i = 0; i < l; i++){ stack.push(pushA[i]); while(!stack.isEmpty() && stack.peek()==pop...
在insert into test(xid) values (1), (3), (5), (8), (11);后,由于xid上是有索引的,该算法总是会去锁住索引记录。现在,该索引可能被锁住的范围如下:(-∞, 1], (1, 3], (3, 5], (5, 8], (8, 11], (11, +∞)。Session A(select * from test where id = 8 for update)执行后...
};//写法2varfindAnagrams =function(s, p) {//res:返回的结果//win:存储窗口中的字符和对应的频次//need:存储需要的异位词的种类和数量//len:need异位词的字符种类//val:滑动窗口中和need中字符数量相同的字符种类constres = [], win = {}, need = {}, pLen = p.length;letlen =0, val =0;fo...
classSolution{public:voidmoveZeroes(vector<int>&nums){intlast_non_zero=0;for(inti=0;i<nums.size();++i){if(nums[i]!=0)nums[last_non_zero++]=nums[i];}for(;last_non_zero<nums.size();last_non_zero++){nums[last_non_zero]=0;}}}; ...
Username for 'https://gitee.com': userName Password for 'https://userName@gitee.com':#私人令牌 分支1 标签0 undefined 贡献代码 同步代码 创建Pull Request 了解更多 对比差异通过 Pull Request 同步 同步更新到分支 通过Pull Request 同步 将会在向当前分支创建一个 Pull ...
949 给定数字能组成的最大时间 - Largest Time for Given Digits C++ Java Python3 Easy 948 令牌放置 - Bag of Tokens C++ Java Python3 Medium 947 移除最多的同行或同列石头 - Most Stones Removed with Same Row or Column C++ Java Python3 Medium 946 验证栈序列 - Validate Stack Sequences C++ Java...
LeetCode_001 "LeetCode 001.两数之和" 示例: 示例代码: 方法一:两层 for 循环暴力破解 测试用例:29个,{(2, 7, 11, 15), tar
583 Delete Operation for Two Strings Solution Medium DP 582 Kill Process Solution Medium Stack 581 Shortest Unsorted Continuous Subarray Solution Easy Array, Sort 576 Out of Boundary Paths Solution Hard DP, DFS 575 Distribute Candies Solution Easy Array 573 Squirrel Simulation Solution Medium Math 572...
比如"abcde" ,"a"、"ab"、"abc"、"abcd"是前缀,而任意连续位置"bc"、"abc"、"cde"等都属于子串!! 1. 2. 3. 4. 今天就是忘了它们的区别,吃了一大亏;当时就想,怎么一简单题还弄出这么多花样。。。 思路:最长公共前缀肯定蕴含在最短子串中,所以for(i from 0 to 最短子串的长度) ,把每个字符串...