題目：傳送門

分析：這個題和62很像，只是最後求得是最短的那條路徑，還是用動態規劃的思想。

$dp\left [ i \right ]\left [ j \right ]= min\left \{ dp\left [ i-1 \right ]\left [ j \right ],dp\left [ i \right ]\left [ j-1 \right ] \right \}+grid[i][j]$

注意第一行和第一列要特殊處理。

#include <algorithm>
#include <string>


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