【LeetCode】#142環形連結串列II(Linked List Cycle II)

題目描述

給定一個連結串列，返回連結串列開始入環的第一個節點。 如果連結串列無環，則返回 null。
為了表示給定連結串列中的環，我們使用整數 pos 來表示連結串列尾連線到連結串列中的位置（索引從 0 開始）。 如果 pos 是 -1，則在該連結串列中沒有環。
說明：不允許修改給定的連結串列。

示例

示例 1：

輸入：head = [3,2,0,-4], pos = 1
輸出：tail connects to node index 1
解釋：連結串列中有一個環，其尾部連線到第二個節點。

示例 2：

輸入：head = [1,2], pos = 0
輸出：tail connects to node index 0
解釋：連結串列中有一個環，其尾部連線到第一個節點。

示例 3：

輸入：head = [1], pos = -1
輸出：no cycle
解釋：連結串列中沒有環。

Description

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

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.

解法

public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        int flag = 0;
        while(fast!=null && fast.next!=null){
            slow = slow.next;
            fast = fast.next.next;
            if(slow==fast){
                flag = 1;
                break;
            }
        }
        if(flag==0){
            return null;
        }
        ListNode node = head;
        while(slow!=node){
            slow = slow.next;
            node = node.next;
        }
        
        return node;
    }
}