【Lintcode】103. Linked List Cycle II

題目地址：

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

給定一個鏈表，判斷其是否有環。如果無環，則返回null，否則返回環的入口。參考https://blog.csdn.net/qq_46105170/article/details/104015181。代碼如下：

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每次走兩步
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                break;
            }
        }
        
        // 如果fast走到了表尾，說明無環，則返回null
        if (fast == null || fast.next == null) {
            return null;
        }
        // 否則新開一個指針從表頭開始走，與slow同時走直到相遇。相遇點即爲所求
        ListNode x = head;
        while (x != slow) {
            x = x.next;
            slow = slow.next;
        }
        return x;
    }
}

時間複雜度$O(n)$，空間$O(1)$。