【Lintcode】103. Linked List Cycle II

题目地址：

https://www.lintcode.com/problem/linked-list-cycle-ii/description

给定一个链表，判断其是否有环。如果无环，则返回null，否则返回环的入口。参考https://blog.csdn.net/qq_46105170/article/details/104015181。代码如下：

public class Solution {
    /**
     * @param head: The first node of linked list.
     * @return: The node where the cycle begins. if there is no cycle, return null
     */
    public ListNode detectCycle(ListNode head) {
        // write your code here
        ListNode slow = head, fast = head;
        while (fast != null && fast.next != null) {
        	// slow每次走一步，fast每次走两步
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                break;
            }
        }
        
        // 如果fast走到了表尾，说明无环，则返回null
        if (fast == null || fast.next == null) {
            return null;
        }
        // 否则新开一个指针从表头开始走，与slow同时走直到相遇。相遇点即为所求
        ListNode x = head;
        while (x != slow) {
            x = x.next;
            slow = slow.next;
        }
        return x;
    }
}

时间复杂度$O(n)$，空间$O(1)$。