摘要： （1）二叉树的中序遍历： 地址链接：http://www.cnblogs.com/aly15109725486/p/7235756.html 代码： public: vector<int> inorderTraversal(TreeNode* root) { vector<int> ret; if( 阅读全文

摘要： 代码： public: vector<int> inorderTraversal(TreeNode* root) { vector<int> ret; if(root==NULL)return ret; inorderHelper(ret,root); return ret; } private: 阅读全文

摘要： 代码： public class Solution { /** * @param grid: a list of lists of integers. * @return: An integer, minimizes the sum of all numbers along its path */ 阅读全文

摘要： 代码： class Solution { public: /** * @param grid: a list of lists of integers. * @return: An integer, minimizes the sum of all numbers along its path */ 阅读全文

摘要： 代码： public class Solution { /** * @param A an integer array * @return a long integer */ public long permutationIndex(int[] A) { // Write your code her 阅读全文

摘要： 代码： class Solution { public: /** * @param matrix: A list of lists of integers * @return: Void */ void setZeroes(vector<vector<int> > &matrix) { // wri 阅读全文

摘要： 代码： class Solution { public: int threeSumClosest(vector<int>& nums, int target) { int closest = nums[0] + nums[1] + nums[2]; int diff = abs(closest - 阅读全文

摘要： 代码： class Solution { public: /** * @param prices: Given an integer array * @return: Maximum profit */ int maxProfit(vector<int> &price) { int re = 0; 阅读全文

摘要： 代码： class Solution { public: /** * Determine whether s3 is formed by interleaving of s1 and s2. * @param s1, s2, s3: As description. * @return: true o 阅读全文

摘要： 代码： void per(vector<int> nums,int start,vector<vector<int> > &result){ if(start==nums.size()-1){ result.push_back(nums); } else{ for(int i=start;i<num 阅读全文