leetcode142 Linked List Cycle II

"""
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.
Note: Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.
"""
"""
判断链表是否有环。有两种做法
第一种是用set()存已经遍历过的结点
如果新的结点在set()里，则返回，有环
"""

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def detectCycle(self, head):
        nodes = set()
        while head:
            if head in nodes:
                return head
            nodes.add(head)
            head = head.next
        return None

"""
解法二：快慢指针
A→B→C       快指针：A, C, B, D 一次走两步
   ↖↓       慢指针：A, B, C, D
    D
根据距离推算：相遇点距环入口的距离 = (头节点距环入口的距离)*（快指针步数-1)
快慢指针相遇在D，让快指针变为head
同步向后，再次相遇即为环的起点
"""

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def detectCycle(self, head):
        if head == None or head.next == None:
            return None
        slow, fast = head, head
        while fast:
            slow = slow.next
            fast = fast.next
            if fast:
                fast = fast.next
                if fast == slow:
                    break
        if slow != fast:
            return None
        fast = head #!!!关键所在
        while slow:
            if fast == slow:
                return fast
            fast = fast.next
            slow = slow.next