64. Minimum Path Sum
这道是非常非常典型的动态规划问题，以前读书的时候记得老师讲的时候死活听不懂，现在工作了几年后，维基百科看了一些介绍，手写一遍AC通过。

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

这道题是先把第一行和第一列的距离算出来，我们从开始位置触发，要么向下，要么向右，因此从起点慢慢从上往下 从左往右计算，最终sum最右下角数据即为结果。
当我洋洋得意的打开讨论区看到大神一篇帖子里，初始版本解法和我上面代码基本一样，难道我还存在更牛的解法？上面程序执行结果：

Runtime: 20 ms, faster than 59.19% of C++ online submissions for Minimum Path Sum.
Memory Usage: 10.5 MB, less than 9.39% of C++ online submissions for Minimum Path Sum.

速度更快解法，相对难懂,gdb跟两编就会有个清晰的理解，相对于方法一少了一个循环：

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

执行结果：

Runtime: 16 ms, faster than 91.86% of C++ online submissions for Minimum Path Sum.
Memory Usage: 10.2 MB, less than 9.39% of C++ online submissions for Minimum Path Sum.

可调试版本程序：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <list>
using namespace std;

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

int main()
{
    vector<vector<int>> path{{1,3,1},{1,5,1},{4,2,1}};
    cout << minPathSum(path);
    return 0;
}