Leetcode 3665. Twisted Mirror Path Count

• [Leetcode 3665. Twisted Mirror Path Count](#Leetcode 3665. Twisted Mirror Path Count)
  • [1. 解题思路](#1. 解题思路)
  • [2. 代码实现](#2. 代码实现)

• 题目链接：3665. Twisted Mirror Path Count

1. 解题思路

这一题相较于上一题就直接很多了，就是一个动态规划，直接写就行了。

2. 代码实现

给出python代码实现如下：

MOD = 10**9+7

class Solution:
    def uniquePaths(self, grid: List[List[int]]) -> int:
        n, m = len(grid), len(grid[0])
        
        DIRECTION = {0: (1, 0), 1: (0, 1)}

        @lru_cache(None)
        def dp(i, j, direction):
            if i == n-1 and j == m-1:
                return 1
            if i < 0 or i >= n or j < 0 or j >= m:
                return 0
            if grid[i][j] == 0:
                return (dp(i+1, j, 0) + dp(i, j+1, 1)) % MOD
            while 0 <= i < n and 0 <= j < m and grid[i][j] == 1:
                direction = 1-direction
                dx, dy = DIRECTION[direction]
                i, j = i+dx, j+dy
            return dp(i, j, direction) % MOD

        return dp(0, 0, 0)

提交代码评测得到：耗时耗时2733ms，占用内存306.37MB。