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.

方法一：
采用递归调用的思想，从左上角开始，每次往右或者下进行移动，到达右下角后，保存结果，然后进行回溯。代码如下，可惜超时：

class Solution {public:    int minPathSum(vector<vector<int>>& grid) {        if(grid.empty() || grid[0].empty())            return 0;        int res = INT_MAX;        m = grid.size();        n = grid[0].size();        dfs(grid, 0, 0, res, 0);        return res;    }    void dfs(vector<vector<int>>& grid, int i, int j, int& res, int tmp) {        tmp += grid[i][j];        if(i == m-1 && j == n-1) {            res = min(res, tmp);        }        if(i < m-1) {            dfs(grid, i+1, j, res, tmp);        }        if(j < n-1) {            dfs(grid, i, j+1, res, tmp);        }    }private:    int m, n;    };

方法二：
采用DP动态规划，定义dp[i][j]数组为从左上角（0， 0）到位置（i, j）的最短路径，则：
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
当i=0，或j=0时，则分别为：
dp[i][j] = dp[i][j-1] + grid[i][j]
dp[i][j] = dp[i-1][j] + grid[i][j]
代码如下：

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