【leetcode-62，63，64 动态规划】不同路径，最小路径和

【【leetcode-62，63，64 动态规划】不同路径，最小路径和】的更多相关文章

[LeetCode] 64. Minimum Path Sum 最小路径和

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. Example: Input: [ [1,3,1], [1,5…

[leetcode]64. Minimum Path Sum最小路径和

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. Input: [ [1,3,1], [1,5,1], [4,2,1…

LeetCode 62，从动态规划想到更好的解法

本文始发于个人公众号:TechFlow,原创不易,求个关注今天是LeetCode专题第36篇文章,我们一起来看下LeetCode的62题,Unique Paths. 题意其实这是一道老掉牙的题目了,我在高中信息竞赛的选拔考试上就见过这题.可想而知它有多古老,或者说多经典吧.一般来说能够流传几十年的算法题,一定是经典中的经典.下面我们就来看下它的题意. 这题的题意很简单,给定一个矩形的迷宫,左上角有一个机器人,右下角是目的地.这个机器人只能向下走或者是向右走,请问这个机器人走到目的地的路径一共…

LeetCode 62. Unique Paths（所有不同的路径）

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…

leetcode@ [62/63] Unique Paths II

class Solution { public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { || obstacleGrid[].size()==) ; ].size(); ][n-]==) ; vector<vector<int> > dp(m); ;i<dp.size();++i) dp[i].resize(n); ;i<dp.size();++i…

LeetCode OJ：Minimum Path Sum（最小路径和）

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. 对m * n 的网格,求其左上角到右下角的最短路径和: DP问题,…

[LeetCode] 62. Unique Paths 唯一路径

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…