[Leetcode] 142. Linked List Cycle II (set / 快慢指针)

142. Linked List Cycle II (题目链接)

Medium

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.

 

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

 

方法1：使用set判重

class Solution:
    def detectCycle(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return None
        d = set([])
        node = head
        while node:
            if node in d:
                return node
            d.add(node)
            node = node.next
        return None

方法2：快慢指针

设置快指针是慢指针速度的两倍，从同一起点出发

设A点为起点，B点为开始出现环的点，C是相遇的点，推导如下

逆时针走：

AB=c, BC=b, CB= a

class Solution:
    def detectCycle(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return None
        slow, fast = head, head
        while fast:
            slow = slow.next
            fast = fast.next.next if fast.next else None
            if slow == fast:
                break
        if slow == fast:
            fast = head
            while fast != slow:
                slow = slow.next
                fast = fast.next
            return fast
        else:
            return None