[LeetCode] 142. 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?

单链表中的环二。题意是给一个链表，如果这个链表中有环，请return环的起点；若没有，return null。找是否有环可以参照[LeetCode] 141. Linked List Cycle的讲解。至于怎么找到环的起点，我这里引用一个非常好的讲解，https://www.cnblogs.com/hiddenfox/p/3408931.html

因为快慢指针的速度是一个2步一个1步，所以当两个指针相遇的时候，fast走过的长度一定是slow的两倍。两者相遇的地方一定是环的起点。至于证明，直接参照引用贴。

时间O(n)

空间O(1)

JavaScript实现

/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var detectCycle = function(head) {
    // corner case
    if (head === null || head.next === null) {
        return null;
    }

    // normal case
    let slow = head;
    let fast = head;
    while (fast !== null && fast.next !== null) {
        slow = slow.next;
        fast = fast.next.next;
        if (fast === slow) {
            let slow2 = head;
            while (slow !== slow2) {
                slow = slow.next;
                slow2 = slow2.next;
            }
            return slow;
        }
    }
    return null;
};

Java实现

/**
 * 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) {
        // corner case
        if (head == null || head.next == null) {
            return null;
        }

        // normal case
        ListNode slow = head;
        ListNode fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                ListNode slow2 = head;
                while (slow != slow2) {
                    slow = slow.next;
                    slow2 = slow2.next;
                }
                return slow;
            }
        }
        return null;
    }
}

LeetCode 题目总结