64. Minimum Path Sum

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.

Example 1:

[[1,3,1],
[1,5,1],
[4,2,1]]

解题思路
该问题是给了一个mxn的二维的数组，要求从数组的左上角到达数组的右下角，找到一个做短的路径。即从左上角开始，每次移动一步，只能向右或者是向下，将每次走到的位置的值加起来，使得到达右下角的时候和最小。我们可以从右下往左上移动，每次计算当前点到右下的最短。f[i, j] = grid[i, j] + min(f[i+1,j], f[i,j+1])，下边和右边两条边线进行特殊处理。
代码展示

#include<iostream>#include<string>#include<math.h>#include<limits>#include<vector>using namespace std;class Solution {public:    int minPathSum(vector<vector<int> >& grid) {        int m = grid.size();        if (m == 0) { return 0; }        int n = grid[0].size();        if (n == 0) { return 0; }        int sol[m+1][n+1];        sol[m-1][n] = 0;        sol[m][n-1] = 0;        int int_max = numeric_limits<int>::max();        for (int i = m-2; i >= 0; i--) {            sol[i][n] = int_max;        }         for (int j = 0; j < n - 1; j++) {            sol[m][j] = int_max;        }         for (int i = m-1; i >= 0; i--) {            for (int j = n-1; j >= 0; j--) {                sol[i][j] = grid[i][j] + min(sol[i+1][j], sol[i][j+1]);            }        }        return sol[0][0];    }};int main(){    int m, n;     cout<<"请输入矩阵的长和宽：";     cin>>m>>n;    vector<vector<int> > grid(m, vector<int>(n));    for(int i=0;i<m;i++){        for(int j=0;j<n;j++){            cin>>grid[i][j];        }    }    Solution solution;    int result=solution.minPathSum(grid) ;    cout<<result<<endl;} 

运行结果展示