Minimum Path Sum

最新推荐文章于 2025-08-21 20:43:22 发布

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本文详细解释了如何在给定的m x n网格中找到从左上角到右下角的最小路径和，只允许向下或向右移动。通过动态规划方法实现，包括初始化、边界处理、核心计算和返回结果等步骤。

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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.

比较简单：

class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) {
        const int m = grid.size();
        const int n = grid[0].size();
        
        int f[m][n];
        f[0][0] = grid[0][0];
        for (int i=1;i<m;i++){
            f[i][0] = f[i-1][0] + grid[i][0];
        }
        for (int i=1;i<n;i++){
            f[0][i] = f[0][i-1] + grid[0][i];
        }
        for (int i=1;i<m;i++){
            for (int j=1;j<n;j++){
                f[i][j] = grid[i][j] + min(f[i-1][j],f[i][j-1]);
            }
        }
        
        return f[m-1][n-1];
    }
};

python

class Solution:
    # @param grid, a list of lists of integers
    # @return an integer
    def minPathSum(self, grid):
        m = len(grid)
        n = len(grid[0])
        
        all = grid[:]
        for i in range(1,n):
            all[0][i] += all[0][i-1]
        
        for i in range(1,m):
            all[i][0] += all[i-1][0]
        
        for i in range(1,m):
            for j in range(1,n):
                all[i][j] += min(all[i-1][j],all[i][j-1])
        
        return all[m-1][n-1]

总结：

1. 使用 const 比较好，传进来的参数不被修改

2. 注意循环的其始