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 1:

[[1,3,1],
[1,5,1],
[4,2,1]]
Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum.

和前两道Unique Paths题的DP思路类似，可写出递推式：

paths[x][y] = grid[x][y] + min(path[x-1][y], path[x][y-1])

代码如下：

class Solution {public:    int minPathSum(vector<vector<int>>& grid) {        int m = grid.size(), n = grid[0].size();        if (m == 0 || n == 0) return 0;        //边界初始化        for (int i = 1; i < m; i++) {            grid[i][0] += grid[i-1][0];        }        for (int i = 1; i < n; i++) {            grid[0][i] += grid[0][i-1];        }        for (int i = 1; i < m; i++) {            for (int j = 1; j < n; j++) {                int min = (grid[i-1][j] > grid[i][j-1]) ? grid[i][j-1] : grid[i-1][j];                grid[i][j] += min;            }        }        return grid[m-1][n-1];    }};