<LeetCode OJ> 64. Minimum Path Sum

64. Minimum Path Sum

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Total Accepted: 62294 Total
Submissions: 183284 Difficulty: Medium

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.

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分析：

很显然的动态规划问题。

令从（1，1）到（m,n）的最小和为result[m,n]

任何一个点都是来自上面的最小和或者来自右边的最小和与当前位置的值相加的加过

显然result[m,n]=min(result[m-1,n]+grid[m,n],result[m,n-1]+grid[m,n])

注意初始化问题

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int row=grid.size();//行
        int col=grid[0].size();
        vector< vector<int> >  result(row);     
        for(int i=0;i <row ;i++)     
            result[i].resize(col,0);//设置数组的大小row行，col列 
        result[0][0]=grid[0][0];//初始化    
        for(int i=1;i<col;i++)//初始化第一行
            result[0][i]=result[0][i-1]+grid[0][i];
        for(int i=1;i<row;i++)//初始化第一列
            result[i][0]=result[i-1][0]+grid[i][0];
        for(int i=1;i<row;i++)//计算中间结果
            for(int j=1;j<col;j++)
                result[i][j]=min(result[i][j-1]+grid[i][j],result[i-1][j]+grid[i][j]);
        return result[row-1][col-1];
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50640213

原作者博客：http://blog.csdn.net/ebowtang