[LeetCode] 141. Linked List Cycle

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?

环形链表。

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 。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/linked-list-cycle
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题意很简单，思路是用快慢指针，慢指针每走一步，快指针走两步。如果快慢指针在某个地方相遇了，说明有环；否则快指针就会遍历到链表尾部从而会退出循环。

时间O(n)

空间O(1)

JavaScript实现

/**
 * @param {ListNode} head
 * @return {boolean}
 */
var hasCycle = function(head) {
    // corner case
    if (head === null || head.next === null) {
        return false;
    }

    // normal case
    let slow = head;
    let fast = head;
    while (fast !== null && fast.next !== null) {
        slow = slow.next;
        fast = fast.next.next;
        if (slow === fast) {
            return true;
        }
    }
    return false;
};

 

Java实现

public class Solution {
    public boolean hasCycle(ListNode head) {
        // corner case
        if (head == null || head.next == null) {
            return false;
        }

        // normal case
        ListNode slow = head;
        ListNode fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                return true;
            }
        }
        return false;
    }
}

 

LeetCode 题目总结