142. Linked List Cycle II

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Total Accepted: 112637
Total Submissions: 363331
Difficulty: Medium
Contributor: LeetCode

Given a linked list, return the node where the cycle begins. If there is no cycle, return null
.
Note: Do not modify the linked list.
Follow up:Can you solve it without using extra space?

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Linked List Two Pointers
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** 解题思路 **
用快慢指针，快指针一次走两步，慢指针一次走一步，来先确定是否有cycle。
如果存在cycle，则将fast = head，再重新 fast =fast.next, slow = slow.next 两个指针都单步走，找到fast = slow点即可确定cycle起始点。

/**
 * 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;
        boolean isCycle = false;
        
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) {
                isCycle = true;
                break; // don't forget to break the loop
            }
        }
        if (!isCycle) return null;
        
        // find the node while cycle begins
        fast = head;
        while (fast != slow) {
            fast = fast.next;
            slow = slow.next;
        }
        
        return slow;
    }
}