LeetCode142. Linked List Cycle II

原题地址：https://leetcode.com/problems/linked-list-cycle-ii/description/

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题目描述：
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

Follow up:
Can you solve it without using extra space?

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题解：首先判断是否有环，思路见上题，在判断有环后，在判断环的入口。
仍然设非环部分长度为a，环部分为b， 慢指针入环后两者相距x，寻找的就是第a+1个节点。
当慢指针走到入口处时，假设快指针已经走了r圈，即2*a = b*r+a+（b-x）=》a = b*r+（b-x） （r >=0）
所以我们再设定两个指针分别指向链表头和快慢指针相遇的位置，当前一个指针走了a步即走到环的入口的时候，第二个指针走了b-x+b*r步，正好也在环的入口，两者相遇。所以要求的就是两者相遇的位置。

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# 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
        """
        a = b = head
        while (a is not None and b is not None):
            a = a.next
            b = b.next
            if b is not None:
                b = b.next
            else:
                return None
            if a == b:
                c = head
                d = a
                while (1):
                    if c == d:
                        return c
                    c = c.next
                    d = d.next
        return None

16 / 16 test cases passed.
Status: Accepted
Runtime: 70 ms