LeetCode 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.

用一个dp二维数组标记当前方格的最小值0.当i == 0 && j == 0的时候，dp[i][j] = grid[i][j];1.对于i==0的时候，为最上面一排，当前方格只能由左边方格来，所以dp[i][j] = dp[i][j-1] + grid[i][j];2.对于j==0的时候，为最左边一排，当前方格只能由上边方格来，所以dp[i][j] = dp[i-1][j] + grid[i][j];3.其他情况下，dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j];最后直到一直递推输出到终点(m-1, n-1)的时候return dp[m-1][n-1];

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