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.

leetcode上做题还是蛮有成就感的。独立思考完成的第三题。解题思路：与unique paths有些类似。建立一个二维数组保存每种grid下的路径的最小和。并且，下一步的最小和依赖于上一步的最小和。得到了递推式：

mins[i][j] = grid[i][j] + min(mins[i-1][j],mins[i][j-1]);

class Solution {public:    int minPathSum(vector<vector<int>>& grid) {               vector<vector<int>> mins(grid.size(), vector<int>(grid[0].size(), 0));        //初始化，没有成功？访问vector越界mins[0][0] = grid[0][0];for (int i = 1; i < grid[0].size(); i++){mins[0][i] = mins[0][i - 1] + grid[0][i];} for (int i = 1; i < grid.size(); i++){mins[i][0] = mins[i - 1][0] + grid[i][0];//vector访问不要越界了}if(grid.size()>1)for (int i = 1; i < grid.size(); i++)for (int j = 1; j < grid[0].size(); j++)//注意访问的起始下表{mins[i][j] = grid[i][j] + min(mins[i-1][j],mins[i][j-1]);}return mins[grid.size()-1][grid[0].size()-1];    }};