小小程序媛  
得之坦然,失之淡然,顺其自然,争其必然

题目

Total Accepted: 47928 Total Submissions: 148011 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.

分析

这道题目是求最小路径和,与LeetCode 62以及LeetCode63本质相同,只需要把存储路径数目的数组改为存储当前最小路径和即可。

AC代码

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        if (grid.empty())
            return 0;

        //求当前矩阵的行数、列数
        int m = grid.size();
        int n = grid[0].size();

        vector<vector<int> > sum(m, vector<int>(n, 0));

        //记录首元素和
        sum[0][0] = grid[0][0];
        int minSum = sum[0][0];

        for (int i = 1; i < m; i++)
        {
            sum[i][0] = sum[i - 1][0] + grid[i][0];
            //此时路径和唯一,也是最小路径和
        }//for

        for (int j = 1; j < n; j++)
        {
            sum[0][j] = sum[0][j - 1] + grid[0][j];
            //此时路径和唯一,也是最小路径和
        }//for

        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];
            }//for
        }//for

        return sum[m - 1][n - 1];
    }
};

GitHub测试程序源码

posted on 2015-09-11 21:44  Coding菌  阅读(95)  评论(0编辑  收藏  举报