64. Minimum Path Sum

Description:

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.

简要题解：

采用动态规划。

子问题c(i, j)： 表示以grid[0][0]为起点，grid[i][j]为终点的最小路径和。

第一个子问题c(0, 0) = grid[0][0]。

子问题c(i , j)与前面的子问题的一般关系：c(i, j) = min(c(i - 1, j), c(i, j - 1)) + grid[i][j]。

代码：

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