pantingting的专栏

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */clas

这里只要对于Vector比较熟悉就很好做了class Solution {public: int removeElement(vector& nums, int val) { vector::iterator it; for(it=nums.begin();it!=nums.end();){ if(*it==val){

class Solution {public: void moveZeroes(vector& nums) { int i=0; int cnt=0; while(i<nums.size()-cnt){ if(nums[i]==0){ for(int j=i+1;j<nums.size

class Solution {public: int searchInsert(vector& nums, int target) { int i; for(i=0;i<nums.size();i++){ if(nums[i]>=target){ break; }

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */clas

递归做法：/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListN

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListNode* in

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.For example:Given n = 13,Return 6, because digit 1 occurred in the followin

Given an integer, write a function to determine if it is a power of two.若一个数字是2的幂，那么它的二进制表示形式里面只有一个1，那么n-1刚好是除了为1的那位之外，其余所有位都为1的数字，举个例子，4的二进制为100，而3 的二进制为011，综上，如果一个数字n&（n-1）==0那么这个数字就是2的幂。class S

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent house

层次遍历我的想法：层序遍历，然后输出每一层最右边的节点代码如下：class Solution {public: vector rightSideView(TreeNode* root) { vector result; vector high; vector low; if(root==NULL) return re

class Solution {public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { ListNode *ans=new ListNode(-1);//哨位节点，尾差； ListNode * tail=ans; while (l1&&l2) { i

class Solution {public: int hash[256]; Solution(){ memset(hash, 0, sizeof(hash)); hash['I']=1; hash['V']=5; hash['X']=10; hash['L']=50; hash['

class Solution {public: string preProcess(string s){//避免回文串的奇偶性讨论 string ans=""; if (s.size()==0) { return "^$"; } ans.append("^"); for (int i=

这道题的Array是已经排序好的。用i，j两个指针，其中i指针以前的数据全部都保持不重复，j指针用于遍历原数组，j指针遇到与前一个数（即i指针所指向的数据）不同时即将数据添加到i指针末尾，并移动i指针。class Solution {public: int removeDuplicates(vector& nums){ int i=0; int j=1;

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)P A H NA P L S I

class Solution {public: vector twoSum(vector& nums, int target) { vectorans,tmp; tmp=nums; int first=0; int second=nums.size()-1; while ((nums[first]+nums[

class Solution {public: char temp[1000];//暂存拆解出来的数字 string str1; string str2; bool isPalindrome(int x) { if (x<0) { return false; } if (x>0) {

String to Integer (atoi)class Solution {public: int myAtoi(string str) { long long ans=0; bool isNegtive=false; //丢弃前面的' ' int validIdx=0; for (int i

class Solution {public: int reverse(int x) { long long ans=0; bool symbol=-1; if (x>0) { symbol=1; } while (x) { ans=ans*10+x%10;//

class NumArray {public: vector accu;//保存以i为结尾得子串的和(包含i)。public: NumArray(vector &nums) {// accu[0]=nums[0]; accu=nums; for (int i =1; i<nums.size(); i++) {

class Solution {public: int hashtable[128]; int lengthOfLongestSubstring(string s) { int ans=0; memset(hashtable,0,sizeof(hashtable));//标记是否被访问过 for (int i=0; i<s.leng

//Definition for singly-linked list.struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} };class Solution {public: ListNode* addTwoNumbers(Lis

class Solution {public: string longestCommonPrefix(vector& strs) { if (strs.size()==0) { return ""; } string ans=strs[0]; for (int i=1; i<strs.size();

Say you have an array for which the ith element is the price of a given stock on day i.If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock),