wishingweed

This is really a classic problem, no more word need for this.class MyQueue { Stack stack1 = new Stack(); Stack stack2 = new Stack(); // Push element x to the back of queue. public void push

public class Solution { public int maxProfit(int[] prices) { if(prices.length == 0) return 0; int min_price = prices[0]; int max_profit = 0; for(int i

this problem is a little hard.we can use the dfs to solve this problem.when you push_back a num and if the sum is

这个题目有几点值得要提的。第一，i＋＋这个表达式的值时i并不是i

I declare two para ; boundary start and boundary end and for each step I

dfs needs more practice.class Solution {public: vector > subsets(vector &S) { vector > answer; vector cur; answer.push_back(cur); if(S.empty()) return answer; sort(S.begin(),S.

we can just use dfs to solve this problem.

class Solution {public:    double pow(double x, int n) {        if(n == 0)            return 1.0;        if(n        {            if(n == INT_MIN)                return 1.0/pow(x,INT_M

这个问题有点困难，参考了网上的代码。。。还需要继续n

DRAW A PICTURE AND Use two pointers.it is ea

/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; *///the proble

pay attention to n and the string.it is easy to understand the code below.class Solution {public: string countAndSay(int n) { // Start typing your C/C++ solution below // DO

Just use the thought of Bubble sort.I think it would have an O(n) algriothm, I would try it later.public class Solution { public void moveZeroes(int[] nums) { for(int j = 0;j < nums.length

map是一个接口而hashmap是它的一种实现方式for  this problem, if you use eclipse,you will find that we need a () after the fucthion "new";another should be pay attention is when I use eclipse as the edcitior

easy problempublic class Solution { public boolean containsDuplicate(int[] nums) { if(nums.length<2) return false; Arrays.sort(nums); for(int i = 1;i<nums.lengt

基本就是遍历的过程中不断的用hash表进行判断，然后不断的维持着这个仍然在范围内的集合。class Solution {public: bool containsNearbyAlmostDuplicate(vector& nums, int k, int t) { map m; int j = 0; for (int i = 0; i

/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Soluti

/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Soluti

Given an index k, return the kth row of the Pascal's triangle.For example, given k = 3,Return [1,3,3,1].Note:Could you optimize your algorithm to use only O(k) extra space?for this

class Solution {public: int minimumTotal(vector > &triangle) { if(triangle.size() == 0) return 0; vector f(triangle.size()); f[0] = triangle[0][0];

转载自felix博客园原题如下Gas StationThere are N gas stations along a circular route, where the amount of gas at station i is gas[i].You have a car with an unlimited gas tank and it costs cost[

about this problem,1.calculate the precise length of the two linked list.2.calculate the difference value of two length of the linked list.then move the head node difference value of steps behin

this algorithm just use two times of binary search, first in the col direction, to

LeetCode OJProblemsPick One!SubmissionsDiscussDonate hqluo11Unique Paths II Total Accepted: 17536 Total Submissions: 62770 My Submissions Question Solution Follow up for "Unique Paths":Now

//注意下，二维数组长度的使用方法public class Solution { public int minPathSum(int[][] grid) { if(grid == null || grid.length == 0) return -1; for(int i = 1;i<grid.length;i++) grid[i][0]

I use iterate method to solve this problem. 1.I use the h

new 和直接定义的区别new在堆上，需要自己释放内存空间。

this problem have two situations.1.when mistake1 is not

this problem just consider whether this matrix is the right matrix.we just consider each col,each row and ea

this problem uses map;1.map anagrams; then we can use anagram[s]

for save space, I write a really ugly code here./** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} *

class Solution {public: string addBinary(string a, string b) { int lengtha = a.length(); int lengthb = b.length(); int length = 0; if(lengthb > lengtha) length = lengtha;

#include class Solution {public: int numTrees(int n) { vector dp(n+1); dp[0] = dp[1] = 1; for(int i=2;i<=n;i++) { dp[i] = 0; for(int j=0;j

这个题目里面出现了一个比较大的问题就是二维的vector没有初始化的时候你如果使用它就会出现

leetcode updates the test set of this problem.so that we must

we ues two pointers to solve this problem.the fast pointer moves two steps each timethe slow pointer moves 1 steps each t

要注意循环链表的情况，如果big在前，small在ho

for this problem , you just need to use two pointers to judge whether it is a palindorome

I use the iterate function to solve this problem.

public class Solution {    public void sortColors(int[] A) {        int r = 0;int b = A.length-1;int temp = 0;        for(int i = 0;i        {            if(A[i]==0 && r                {