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 = 0
        while i < n and (i < 2 or distance[i] > distance[i - 2]):
            i += 1

        if 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 += 1

        return i != n