leetcode [142]Linked List Cycle II-优快云博客

本文介绍了一种高效检测链表中环入口的算法。通过快慢指针判断链表是否存在环，计算环的长度，并最终定位到环的入口节点。文章提供了详细的解题思路及Java实现代码。

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?

题目大意：

如果链表中存在环，返回环的入口，否则返回空。

解法：

1.先使用快慢指针，快指针走两步，慢指针走一步，判断链表中是否存在环，如果快慢指针相遇，则不存在，否则快指针先走到链表最后，返回null。

2.然后再让快慢指针相遇一次，计算得到环的长度

3.重新让快慢指针都指向头结点，快指针先走环的长度的步数，当快慢指针再相遇的时候，该节点就是环的入口。

java:

public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode slow=head,fast=head;
        if(head==null) return null;
        while(fast!=null&&fast.next!=null){
            slow=slow.next;
            fast=fast.next.next;
            if(slow==fast) break;
        }
        if(fast==null || fast.next==null) return null;
        int count=0;
        while(count==0||slow!=fast){
            slow=slow.next;
            fast=fast.next.next;
            count++;
        }
        slow=head;
        fast=head;
        while(count!=0){
            fast=fast.next;
            count--;
        }
        while(slow!=fast){
            slow=slow.next;
            fast=fast.next;
        }
        return slow;
    }
}

转载于:https://www.cnblogs.com/xiaobaituyun/p/10734628.html