Two Sum(Medium)

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[array] leetcode - 39. Combination Sum - Medium

leetcode - 39. Combination Sum - Medium descrition Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number may be chosen from…

【leetcode】Binary Tree Maximum Path Sum (medium)

Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. 找树的最大路径和注意路径可以从任意点起始和结束. 我发现我真的还挺擅长树的题目的,递归不难.就是因为有个需要比较的量(最大和),所以需要再写一个函数. 因为路径可以从任意点起始和结束,所以每次递归的时候左右子树小于等于0的就可以不管了. #include <iostream> #include…

39. Combination Sum(medium, backtrack 的经典应用, 重要)

Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number may be chosen from C unlimited number of times. Note: All numbers (in…

[leetcode] 64. Minimum Path Sum (medium)

原题简单动态规划重点是:grid[i][j] += min(grid[i][j - 1], grid[i - 1][j]); class Solution { public: int minPathSum(vector<vector<int>> &grid) { for (int i = 1; i < grid.size(); i++) { grid[i][0] += grid[i - 1][0]; } for (int i = 0; i < grid.si…

1.遍历一次nums,利用map记录已经遍历过的nums[i],并记录其index :i+1 2.每次求出b=target-nums[i],检测b是否在map中,如果在,则已经得到答案,否则,把nums[i]存到map里面 AC代码: class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int,int> exist; vector&l…

[array] leetcode - 40. Combination Sum II - Medium

leetcode - 40. Combination Sum II - Medium descrition Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Each number in C may only be used once in the combinat…

【目录】Leetcode

Leetcode 1.动态规划 Palindrome Partitioning II(hard) ☆ Distinct Subsequences(hard) Edit Distance (hard) Interleaving String (hard) Regular Expression Matching (hard) ★ Wildcard Matching(hard) ★ 大神太牛了 Longest Valid Parentheses (hard)★ Scramble String (har…