Python | Leetcode Python题解之第335题路径交叉

题目：

题解：

class Solution:def isSelfCrossing(self, distance: List[int]) -> bool:n = len(distance)# 处理第 1 种情况i = 0while i < n and (i < 2 or distance[i] > distance[i - 2]):i += 1if i == n:return False# 处理第 j 次移动的情况if ((i == 3 and distance[i] == distance[i - 2])or (i >= 4 and distance[i] >= distance[i - 2] - distance[i - 4])):distance[i - 1] -= distance[i - 3]i += 1# 处理第 2 种情况while i < n and distance[i] < distance[i - 2]:i += 1return i != n