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.

分析

可以延续第62、63两道题的思想，从后往前逆推。不同的是，这次要取其后面可能路径中最小的值，并且加上自身的值。

实现

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        if(grid.empty()||grid[0].empty()) return 0;
        int m=grid.size(), n=grid[0].size();
        int dp[m][n];
        dp[m-1][n-1] = grid[m-1][n-1];
        for(int i=m-2; i>=0; i--)
            dp[i][n-1] = dp[i+1][n-1] + grid[i][n-1];
        for(int i=n-2; i>=0; i--)
            dp[m-1][i] = dp[m-1][i+1] + grid[m-1][i];
        for(int i=m-2; i>=0; i--)
            for(int j=n-2; j>=0; j--)
                dp[i][j] = min(dp[i][j+1], dp[i+1][j]) + grid[i][j];
        return dp[0][0];
    }
};

思考

感觉这种类型的题基本没问题了呀（flag已立……）。