标签：
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.
Array Dynamic Programming
这题是一个很简单的动态规划题
#include<iostream>
#include<vector>
using namespace std;
int minPathSum(vector<vector<int>>& grid) {
int m=grid.size();
int n=grid[0].size();
int **ary=new int*[m];
for(int i=0;i<m;i++)
ary[i]=new int[n];
ary[m-1][n-1]=grid[m-1][n-1];
for(int i=n-2;i>=0;i--)
ary[m-1][i]=grid[m-1][i]+ary[m-1][i+1];
for(int i=m-2;i>=0;i--)
ary[i][n-1]=grid[i][n-1]+ary[i+1][n-1];
for(int i=m-2;i>=0;i--)
for(int j=n-2;j>=0;j--)
{
ary[i][j+1]<ary[i+1][j]?ary[i][j]=ary[i][j+1]:ary[i][j]=ary[i+1][j];
ary[i][j]=ary[i][j]+grid[i][j];
}
int a=ary[0][0];
for(int i=0;i<m;i++)
delete []ary[i];
delete []ary;
return a;
}
int main()
{
}
leetcode_64题——Minimum Path Sum（动态规划）
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原文地址：http://www.cnblogs.com/yanliang12138/p/4560145.html
