Minimum Path Sum

一、题目

64. Minimum Path Sum

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.

该题源地址为https://leetcode.com/problems/minimum-path-sum/description/

二、我的解法

class Solution {public:    int minPathSum(vector<vector<int>>& grid) {        int rowlen = grid.size();        int collen = grid[0].size();                vector<vector<int> > dist(rowlen);                for (int i = 0; i < rowlen; i++) {            dist[i].resize(collen);        }        dist[0][0] = grid[0][0];                for (int i = 1; i < rowlen; i++) {            dist[i][0] = dist[i - 1][0] + grid[i][0];        }            for (int j = 1; j < collen; j++) {            dist[0][j] = dist[0][j - 1] + grid[0][j];        }                for (int i = 1; i < rowlen; i++) {            for (int j = 1; j < collen; j++) {                dist[i][j] = min(dist[i][j - 1] , dist[i - 1][j]) + grid[i][j];               }        }        return dist[rowlen - 1][collen - 1];    }};

三、简要分析

这道题可以构建一个数组，每一项的值代表左上角到（i,j）这个格子的最短路径，

因为只能往下和往右走，所以左上角到格子（i，j）的最短路径为
min(dist[i][j - 1] , dist[i - 1][j]) + grid[i][j]，所以只要
知道格子（i，j）的左边和上边格子的dist值，就可以求出格子（i，j）的dist值，
这样，扫描一遍数组就行