LeetCode 142. Linked List Cycle II

题目

思路

快慢指针。先用快慢指针找到是否有环。当有环时，慢指针设置为head，快指针步长设置为1，继续遍历，相遇的点就是入口。

代码

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def detectCycle(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        p1 = head; p2 = head
        while p1 and p2 and p2.next:
            p1 = p1.next
            p2 = p2.next.next
            if p1 == p2:
                p1 = head
                while p1 and p2:
                    if p1 == p2:
                        return p1
                    p1 = p1.next
                    p2 = p2.next
        return None