LeetCode算法笔记--环形链表II

算法题热题HOT100 快慢指针

字数统计: 398阅读时长: 1 min

 2021/01/04   Share

LeetCode算法笔记-Day58

142. Linked List Cycle 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.

Example1:

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.

Example2:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

方法一：Floyd 算法（快慢指针）

Analyse:

类似“追及问题”
两个人在环形跑道上赛跑，同一个起点出发，一个跑得快一个跑得慢，在某一时刻，跑得快的必定会追上跑得慢的，只要是跑道是环形的，不是环形就肯定追不上。
给定找到的相遇点，初始化额外的两个指针： ptr1 ，指向链表的头， ptr2 指向相遇点。然后，我们每次将它们往前移动一步，直到它们相遇，它们相遇的点就是环的入口，返回这个节点。

Answer：

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var detectCycle = function(head) {
    let _slow = head,
        _fast = head,
        _pre = head,
        _meet;

    // 确定是否有环，确定快慢指针相遇点
    while(_fast){
        if(_fast.next == null) return null;
        _fast = _fast.next.next;
        _slow = _slow.next;
        if(_fast == _slow){
            _meet = _fast;
            break;
        }
    }

    // 确定环入口
    if(_fast){
        while(true){
            if(_pre == _meet) return _pre;
            _pre = _pre.next;
            _meet = _meet.next;
        }
    }else{
        return null;
    }
    
};

时间复杂度：O(n)
空间复杂度：O(n)

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CATALOG

1. LeetCode算法笔记-Day58
1. 1.1. 142. Linked List Cycle II
2. 1.2. 方法一：Floyd 算法（快慢指针）
  1. 1.2.1. Analyse:
  2. 1.2.2. Answer：

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