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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.

解题思路：

求一个矩形方格，方格中填充有非负数，由方格左上角向右下角走，只能向右或者向下走，经过的方格数相加，得出的最小和是多少。

利用动态规划，我们可以将方格中的数字改为到该方格的最小和，则grid[i][j]=min{grid[i][j]+grid[i-1][j],grid[i][j]+grid[i][j-1]},其中i>0,j>0.特殊情况为：grid[0][0]不变，第一行为左边方格值+本方格值。第一列为上方方格值+本方格值。

代码如下：

<pre name="code" class="cpp">#include <iostream>
#include <vector>
using namespace std;
int minPathSum(vector<vector<int>>& grid) ;
void main()
{
	vector<vector<int>> grid;
	int a[5]={1,3,1,0,2};
	vector<int> temp(a,a+5);
	grid.push_back(temp);
	int b[5]={4,3,7,12,2};
	vector<int> temp1(b,b+5);
	grid.push_back(temp1);
	int c[5]={9,3,1,5,2};
	vector<int> temp2(c,c+5);
	grid.push_back(temp2);
	for (int i=0;i<grid.size();i++)
	{
		for (int j=0;j<5;j++)
		{
			cout<<grid[i][j]<<" ";
		}
		cout<<endl;
	}
	cout<<"The minPathSum is: ";
	cout<<minPathSum(grid);
	cout<<endl;
	cout<<"After minPathSum function: "<<endl;
	for (int i=0;i<grid.size();i++)
	{
		for (int j=0;j<5;j++)
		{
			cout<<grid[i][j]<<" ";
		}
		cout<<endl;
	}

}
int minPathSum(vector<vector<int>>& grid) {
	vector<int> temp;
	for (int i=0;i<grid.size();i++)
	{
		 temp=grid[i];
		for (int j=0;j<temp.size();j++)
		{
			if (i==0&&j==0)
			{
				continue;
			}
			else if (i==0&&j>0)
			{
				grid[i][j]+=grid[i][j-1];
			}
			else if (j==0&&i>0)
			{
				grid[i][j]+=grid[i-1][j];
			}
			else
			{
				int min;
				if (grid[i-1][j]>grid[i][j-1])
				{
					min=grid[i][j-1];
				}
				else
					min=grid[i-1][j];
				grid[i][j]+=min;
			}
		}
	}
	return grid[grid.size()-1][temp.size()-1];

}

leetcode---Minimum Path Sum

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原文地址：http://blog.csdn.net/sinat_24520925/article/details/45872619