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.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

求从矩阵的左上角到右下角的最小路径和，每次只能向右和向下移动。

C++:

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