0141.LinkedListCycle.js

// 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?

// 来源：力扣（LeetCode）
// 链接：https://leetcode-cn.com/problems/linked-list-cycle
// 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

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

// 🎨 方法一：快慢指针

// 📝 思路：用快慢指针遍历链表，存在环必定会相遇

/**
 * @param {ListNode} head
 * @return {boolean}
 */
var hasCycle = function (head) {
    let slow = fast = head;

    while (slow && fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;

        if (slow === fast) {
            return true
        }
    }

    return false
};