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https://github.com/SunnyQjm/algorithm-review.git
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59 lines
1.9 KiB
Python
59 lines
1.9 KiB
Python
#!/usr/bin/env python
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# coding=utf-8
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#######################################################################################
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# Leetcode 64 最小路径和
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#
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# 给定一个包含非负整数的 m x n 网格，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。
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# 说明：每次只能向下或者向右移动一步。
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#
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# 示例:
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# 输入:
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# [
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# [1,3,1],
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# [1,5,1],
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# [4,2,1]
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# ]
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# 输出: 7
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# 解释: 因为路径 1->3->1->1->1 的总和最小。
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#######################################################################################
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class Solution:
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def minPathSum(self, grid):
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"""
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:type grid: List[List[int]]
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:rtype int
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(knowledge)
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思路：
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1. 采用动态规划；
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2. dp[i][j] => 表示第i + 1行第j + 1列的网格到右下角网格的最短路径开销(本题可以直接复用grid作为dp数组)
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3. 状态转移方程：
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f(i, j) = grid[i][j] i = m-1 && j = n - 1
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grid[i][j] + f(i + 1, j) i < m - 1 && j = n - 1
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grid[i][j] + f(i, j + 1) i = m - 1 && j < n - 1
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grid[i][j] + min{f(i + 1, j), f(i, j + 1)} i < m - 1 && j < n - 1
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"""
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m, n = len(grid), len(grid[0])
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# 先处理最右侧一排
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for i in range(m - 2, -1, -1):
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grid[i][n - 1] += grid[i + 1][n - 1]
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# 先处理最底下一行
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for j in range(n - 2, -1, -1):
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grid[m - 1][j] += grid[m - 1][j + 1]
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for i in range(m - 2, -1, -1):
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for j in range(n - 2, -1, -1):
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grid[i][j] += min(grid[i + 1][j], grid[i][j + 1])
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return grid[0][0]
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if __name__ == '__main__':
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solution = Solution()
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print(solution.minPathSum([[1, 3, 1], [1, 5, 1], [4, 2, 1]]), "= 7")
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