leetcode-142. Linked List Cycle II 环形链表 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?

给定一个链表,返回链表开始入环的第一个节点。 如果链表无环,则返回 null

为了表示给定链表中的环,我们使用整数 pos 来表示链表尾连接到链表中的位置(索引从 0 开始)。 如果 pos 是 -1,则在该链表中没有环。

说明:不允许修改给定的链表。

示例 1:

输入:head = [3,2,0,-4], pos = 1
输出:tail connects to node index 1
解释:链表中有一个环,其尾部连接到第二个节点。

示例 2:

输入:head = [1,2], pos = 0
输出:tail connects to node index 0
解释:链表中有一个环,其尾部连接到第一个节点。

示例 3:

输入:head = [1], pos = -1
输出:no cycle
解释:链表中没有环。

进阶:
你是否可以不用额外空间解决此题?

思路:和上一题一样

第一种:哈希表,所有元素放入哈希表,最后一个元素下一个元素在哈希表中,就返回 哈希表元素,缺点是空间 大

第二种:快慢指针,slow走一步,quick 走两步,若quick或者quick->next为空则不存在环,如果碰头,就是有环,然后 一个指针从环这里,另一个指针从head开始,每个 走一步,知道最终 两个相遇就是要的交点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode *quick=head,*slow=head;
        while(quick && quick->next)
        {
            slow=slow->next;
            quick=quick->next->next;
            if(slow == quick) break;
        }
        if(!quick || !quick->next) return NULL;
        slow = head;
        while(slow!=quick)
        {
            slow=slow->next;
            quick=quick->next;
        }return quick;
    }
};

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转载自blog.csdn.net/qq_21997625/article/details/86546962