题目：原题链接（中等）
标签：数组、双指针
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N ) O(N) O(N) | O ( N ) O(N) O(N) | 460ms (33.64%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class Solution:
def circularArrayLoop(self, nums: List[int]) -> bool:
visited = set()
for i1, num in enumerate(nums):
if i1 in visited:
continue
flag1 = 1 if nums[i1] > 0 else -1
circle = set()
i2 = i1
while i2 not in circle:
flag2 = 1 if nums[i2] > 0 else -1
if flag1 != flag2:
break
circle.add(i2)
i2 = (i2 + nums[i2]) % len(nums)
else:
if len(circle) != 1 and i2 != (i2 + nums[i2]) % len(nums): # 进入单个数字的循环
return True
visited &= circle
return False
