leetcode 141. Linked List Cycle

最新推荐文章于 2024-01-11 13:46:47 发布

转载最新推荐文章于 2024-01-11 13:46:47 发布 · 75 阅读

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原文链接：http://www.cnblogs.com/jamieliu/p/10351186.html

本文介绍两种检测链表中是否存在循环的有效方法。一是使用哈希表，时间复杂度为O(n)，空间复杂度为O(n)；二是采用双指针技巧，即快慢指针法，时间复杂度为O(n)，但仅需常数级空间O(1)。通过实例展示了不同情况下链表循环的判断。

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

To represent a cycle in the given linked list, we use an integer poswhich 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.

Example 1:

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

Example 2:

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

Example 3:

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

Follow up:

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

解法一：哈希表

时间 O（n）空间 O（n）

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        Map<ListNode,Integer> map = new HashMap<>();
        int i=0;
        while(head != null){
            if(map.containsKey(head)) return true;
            map.put(head,i);
            head = head.next;
            i++;
        }
        return false;
    }
}

解法二：很巧妙的用两个指针，一快一慢。

runner每次跨两步，walker每次跨一步。如果有circle的话一定会有一天runner追上walker

时间 O（n）空间 O（1）

public class Solution {
    public boolean hasCycle(ListNode head) {
        if(head == null) return false;
        ListNode walker = head;
        ListNode runner = head.next;
        while(walker != runner) {
            if(runner ==null || runner.next == null){
                return false;
            }
            walker = walker.next;
            runner = runner.next.next;
        }
        return true;
    }
}

转载于:https://www.cnblogs.com/jamieliu/p/10351186.html