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✅141. 环形链表 #78

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bazinga-web opened this issue Aug 2, 2020 · 2 comments
Open

✅141. 环形链表 #78

bazinga-web opened this issue Aug 2, 2020 · 2 comments

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@bazinga-web
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141.环形链表

给定一个链表,判断链表中是否有环。

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。

示例 1:

输入:head = [3,2,0,-4], pos = 1
输出:true
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:true
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:false
解释:链表中没有环。
``` 
**进阶:**

> 你能用 O(1)(即,常量)内存解决此问题吗?
@bazinga-web
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bazinga-web commented Aug 2, 2020

解题思路:定义快慢两个指针,如果成环的话一定会在某个节点相遇

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

/**
 * @param {ListNode} head
 * @return {boolean}
 */
var hasCycle = function(head) {
    if (head === null) return false;
    let slow = head;
    let fast = head;
    while (fast.next !== null &&  fast.next.next !== null) {
        slow = slow.next;
        fast = fast.next.next;
        if (slow === fast) return true;
    }
    return false;
};

@Ray-56
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Ray-56 commented Aug 3, 2020

快慢指针

  • 只要是有环形,一定会有相等,所以只需要让一直执行
  • 没有环形,则next存在null遍历完毕直接返回false
var hasCycle = function(head) {
    if (!head) return false;
    let first = head;
    let slow = head;
    while (first.next && first.next.next) {
        slow = slow.next;
        first = first.next.next;
        if (slow === first) return true;
    }
    return false;
};

@Ray-56 Ray-56 changed the title 141. 环形链表 ✅141. 环形链表 Aug 3, 2020
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