Minimum Path Sum问题及解法

问题描述：

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.

问题分析：

分析同Unique Paths。构造状态转移的dp数组，确定关系dp[i][j] = min(dp[i - 1][j],dp[i][j - 1]) + grid[i - 1][j - 1]，最终得到答案dp[m][n]。

过程详见代码：

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