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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

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

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

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

题目链接 有点动态规划的味道。 #include int main() { char s[100]; int p=0, pa=0, pat=0; scanf("%s", s); int len = strlen(s); for (int i = 0; i { switch (s[i]) { case 'p': p += 1; break; case 'a':

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) {        

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 =

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

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

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 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

题目： 把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。 输入一个非递减排序的数组的一个旋转，输出旋转数组的最小元素。 例如数组{3,4,5,1,2}为{1,2,3,4,5}的一个旋转，该数组的最小值为1。 NOTE：给出的所有元素都大于0，若数组大小为0，请返回0。 旋转之后的数组实际上可以划分成两个有序的子数组：前面子数组的大小都大于后面子数组中的

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

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

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 X X O O X X X O X X

自以为对二进制有很深的理解，面对这种水题，切了好久。。。 题目链接 #include #include typedef long long  ll; bool compare(ll a, ll b, ll c)//防止大数相加溢出 { if (a > 0 && b > 0) { if (a > LLONG_MAX - b) return true; } if

#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

题目链接 int main() { char a[100], b[100]; int a1, b1, c1, a2, b2, c2,a3,b3,c3; scanf("%d.%d.%d", &a1, &b1, &c1); scanf("%d.%d.%d", &a2, &b2, &c2); int ca1, ca2;//表示进位 c3 = (c1 + c2) % 29; ca1 =

点击打开链接 老TLE，可以看出浮点运算的复杂程度远大于整数运算。怎么优化。。。 int main() { float f1, f2; int dignum; scanf("%d", &dignum); scanf("%f %f", &f1, &f2); float t1 = f1, t2 = f2; int count1 = 0, count2 = 0; while (t

counting the pat int main() { int p=0, pa=0, pat = 0; char s[100000] = { 0 }; scanf("%s", s); int i = 0; while (s[i]!=0) { switch (s[i]) { case 'P': p++; break; case 'A': pa += p; pa

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; }

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  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

#include<iostream>#include<vector>#include<stack>#include<cstdlib>#include<algorithm>using namespace std; /**把数组分为两部分，轴pivot左边的部分都小于轴右边的部分**/template <typename Compara...