141. Linked List Cycle - 环形链表

Tags - 题目标签

Description - 题目描述

EN:

Given head, the head of a linked list, determine if the linked list has a cycle in it.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter.

Return true if there is a cycle in the linked list. Otherwise, return false.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.

Example 3:

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

Constraints:

The number of the nodes in the list is in the range [0, 10⁴].
-10⁵ <= Node.val <= 10⁵
pos is -1 or a valid index in the linked-list.

Follow up: Can you solve it using O(1) (i.e. constant) memory?

ZH-CN:

给你一个链表的头节点 head ，判断链表中是否有环。

如果链表中有某个节点，可以通过连续跟踪 next 指针再次到达，则链表中存在环。为了表示给定链表中的环，评测系统内部使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。注意：pos 不作为参数进行传递 。仅仅是为了标识链表的实际情况。

如果链表中存在环 ，则返回 true 。否则，返回 false 。

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。

提示：

链表中节点的数目范围是 [0, 10⁴]
-10⁵ <= Node.val <= 10⁵
pos 为 -1 或者链表中的一个 有效索引 。

进阶：你能用 O(1)（即，常量）内存解决此问题吗？

Link - 题目链接

LeetCode - LeetCode-CN

Latest Accepted Submissions - 最近一次 AC 的提交

Language	Runtime	Memory	Submission Time
typescript	80 ms	40.5 MB	2021/11/04 21:13

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function hasCycle(head: ListNode | null): boolean {
  if (!head) {
    return false;
  }
  let slow = head, fast = head;
  while(fast) {
    slow = slow.next ?? null;
    fast = fast.next?.next ?? null;
    if (slow === fast && fast) {
      return true;
    }
  }
  return false;
};

My Notes - 我的笔记

慎用 optional chainning 来写算法题，如果非要用的话注意判空条件。不用 option chainning:

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

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

用了 option chainning:

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function hasCycle(head: ListNode | null): boolean {
  if (!head) {
    return false;
  }
  let slow = head, fast = head;
  while(fast) {
    slow = slow.next ?? null;
    fast = fast.next?.next ?? null;
    if (slow === fast && fast) {
      return true;
    }
  }
  return false;
};

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