fly away!

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.Return a deep copy of the list. struct RandomListNode {     i

题目描述Given a binary tree, find the maximum path sum.The path may start and end at any node in the tree.For example:Given the below binary tree, 1 / \ 2 3

题目描述Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start toend, such that:Only one letter can be changed at a timeEach inte

Given a string s, partition s such that every substring of the partition is a palindrome.Return all possible palindrome partitioning of s.For example, given s ="aab",Return [ ["aa","b

好久不用Java来写了，花了好久才accept题目描述输入一颗二叉树和一个整数，打印出二叉树中结点值的和为输入整数的所有路径。路径定义为从树的根结点开始往下一直到叶结点所经过的结点形成一条路径。public class TreeNode {    int val = 0;    TreeNode left = null;    TreeNode right =

期末爆炸，昨晚作业来写一道题~~~题目描述输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历的结果。如果是则输出Yes,否则输出No。假设输入的数组的任意两个数字都互不相同。public class Solution {    public boolean VerifySquenceOfBST(int [] sequence) {       

题目链接参考了大神的代码#include #include class Solution {public:void gen_path(unordered_map>& father, vector& path, string start, string word,vector>& ret){path.push_back(word);if (start == w

本以为暴力枚举会超时，但竟然过了。。。#include #include #include int longestPalindrome(string s) {int left = 0, right = s.length() - 1;//为奇数时int len1 = 0;int len2 = 0;int start1 = left;//标记起始点int sta

点击打开链接class Solution {public:bool solve(string s1, string s2, int left1, int right1, int left2)//串的位置{int dist = right1 - left1;if (dist == 1){return s1[left1] == s2[left2];}int la

点击打开链接class Solution {public:    bool isdigit(char c){return c >= '0'&&c }int numDecodings(string s) {int dp[1000] = { 1 };//dp[i]表示前i个字符的方        if (s.length() == 0 || s[0] == '0')

点击打开链接 struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}  };class Solution {public:   vector

题目链接Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.For example,Given:s1 ="aabcc",s2 ="dbbca",When s3 ="aadbbcbcac", return true.When s3 ="aadbbbac

题目链接 struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };class Solution {public:    vector

点击打开链接 struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };class Solution {public:    TreeN

题目链接class Solution {public:int a[100];bool issuit(int a[], int n, int row, int col)//判断是否适合放置{for (int i = 0; i {if (abs(a[i] - col) == abs(row - i) || col == a[i])return false;}

#include #include #include #include #include using namespace std;class st{private:vector vec;public:st(){};~st(){};void pop(){if (vec.size() == 0){printf("%s\n", "Inv

Given a 2D board containing'X'and'O', capture all regions surrounded by'X'.A region is captured by flipping all'O's into'X's in that surrounded region .For example,X X X XX O O XX X O XX

题目描述Follow up for problem "Populating Next Right Pointers in Each Node".What if the given tree could be any binary tree? Would your previous solution still work?Note:You may on