题目:
题解:
python
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