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LinkedList - 142. Linked List Cycle II

LinkedList - 142. Linked List Cycle II

  1. 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.

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

思路:

题目是让找到环的起点,做法就是用快慢指针,开始的时候,两个指针都从head出发,然后快指针走两步,慢指针一次走一步,相遇就说明有环,这个时候,再用慢指针从链表头开始走,快指针变成走一步,这样两个指针再次相遇的地方就是环的入口。

详细解释一下推算过程:
假设链表头到环到起点距离是L, 环的长度是S, 快指针一次走两步, 走的长度是2F,慢指针一次走一步,相遇的时候走的长度是F,假设第一次相遇地点距离环起点是U,相遇时候,快指针在环内走了k圈。
那么
 L + k * S + (S - U) = 2(L + (S - U))
也就是
 L = (k - 1) * S + U, (U <= S)
所以存在一个点,能够让相同速度的两个指针各自走完L和U,然后刚好相遇。

代码:

java:

/**

 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        if (head == null || head.next == null) return null;

        ListNode slow = head;
        ListNode fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (fast == slow) {
                while (head != fast) {
                    fast = fast.next;
                    head = head.next;
                }
                return head;
            }
        }
        
        return null;
    }
}

go:

/**

 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func detectCycle(head *ListNode) *ListNode {
    var (
        res *ListNode
        slow = head
        fast = head
    )

    if head == nil || head.Next == nil {
        return res
    }

    for fast != nil && fast.Next != nil {
        slow = slow.Next
        fast = fast.Next.Next
        if slow == fast {
            fast = head
            for slow != fast {
                slow = slow.Next
                fast = fast.Next
            }
            return slow
        }
    }

    return res
}


标题:LinkedList - 142. Linked List Cycle II
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/07/29/1564413775423.html

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