64. Minimum Path Sum

最新推荐文章于 2022-08-02 13:05:05 发布

无名_1989

最新推荐文章于 2022-08-02 13:05:05 发布

阅读量91

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/dongyu_1989/article/details/80013458

本文介绍了一个算法问题：在一个填充了非负数的矩阵中找到从左上角到右下角的路径，使得路径上的数字之和最小。该文详细解释了如何通过动态规划的方法解决这一问题，并给出了一段具体的C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        
        int m = grid.size();
        int n = grid[0].size(); 
        vector<vector<int> > sum(m, vector<int>(n, grid[0][0]));
        for(int i=1; i<grid[0].size(); i++)
        {
            sum[0][i] = sum[0][i-1] + grid[0][i];
        }
        
        for(int j=1; j<grid.size(); j++)
            sum[j][0] = sum[j-1][0] + grid[j][0];
        
        for(int i=1; i<grid.size(); i++)
        {
            for(int j=1; j<grid[0].size(); j++)
            {
                if(sum[i-1][j]<sum[i][j-1])
                {
                    sum[i][j] = sum[i-1][j] + grid[i][j];
                    
                }else
                {
                    sum[i][j] = sum[i][j-1] + grid[i][j];
                }
               cout<<i<<":"<<j<<":"<<sum[i][j]<<endl;
            }
        }
        return sum[grid.size()-1][grid[0].size()-1];
        
    }
};