LeetCode||64. Minimum Path Sum

原创于 2017-11-21 14:28:46 发布 · 282 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #python

leetcode 同时被 2 个专栏收录

99 篇文章

订阅专栏

python

69 篇文章

订阅专栏

本文探讨了一个经典的动态规划问题——给定一个填充了非负数的网格，如何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。文章详细解释了解决方案，并通过一个实例展示了如何计算最小路径和。

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 1:

[[1,3,1],
 [1,5,1],
 [4,2,1]]

Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum.

这题还是类似的动态规划，判断条件变为取两个数之间最小的。

因此，例子中的结果矩阵为：

[[1,4,5],
 [2,7,6],
 [6,8,7]]

所以结果为7

class Solution(object):
    def minPathSum(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        row = len(grid)
        col = len(grid[0])
        res = [[0 for i in range(col)] for j in range(row)]
        res[0][0] = grid[0][0]
        for i in range(1, row):
        	res[i][0] = res[i-1][0] + grid[i][0]
        for i in range(1, col):
        	res[0][i] = res[0][i-1] + grid[0][i]
        for i in range(1, row):
        	for j in range(1, col):
        		res[i][j] = min(res[i-1][j], res[i][j-1]) + grid[i][j]
        return res[-1][-1]