leetcode142. 环形链表 II

leetcode142. 环形链表 II

题目

思路

集合法

• 将节点存入set，若重复出现则说明是环

快慢指针法

• 分别定义 fast 和 slow 指针，从头结点出发，fast指针每次移动两个节点，slow指针每次移动一个节点，如果 fast 和 slow指针在途中相遇 ，说明这个链表有环。
• 初次相遇后，将slow设为头结点，slow和fast这两个指针每次只走一个节点， 当这两个指针相遇的时候就是环形入口的节点。

代码

集合法

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        node_set = set()
        current = head
        while current:
            if current in node_set:
                return current
            else:
                node_set.add(current)
                current = current.next
        return None

快慢指针法

class Solution:
    def detectCycle(self, head: ListNode) -> ListNode:
        slow = head
        fast = head
        
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            
            # If there is a cycle, the slow and fast pointers will eventually meet
            if slow == fast:
                # Move one of the pointers back to the start of the list
                slow = head
                while slow != fast:
                    slow = slow.next
                    fast = fast.next
                return slow
        # If there is no cycle, return None
        return None