64. Minimum Path Sum

题目

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.

思路

本题使用递归的思路，维护一个二位数组res[m][n]存储到达当前结点最小和；状态转移方程为：res[i][j] = grid[i][j] + min(res[i-1][j],res[i][j-1])

代码

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