141. Linked List Cycle

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本文介绍两种有效的方法来判断链表中是否存在循环：哈希表法和双指针法。哈希表法通过记录已访问过的节点，实现O(n)的时间复杂度和O(n)的空间复杂度；双指针法则利用快慢指针，达到O(n)的时间复杂度和O(1)的空间复杂度。

题目描述

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 pos which 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.

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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.
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Example 3:

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

Approach1:Hash Table
Complexity analysis

Time complexity : O(n). 
We visit each of the nnn elements in the list at most once. 
Adding a node to the hash table costs only O(1)time.

Space complexity: O(n). 
The space depends on the number of elements added to the hash table,
 which contains at most nnn elements.

class Solution{
    public boolean hasCycle(ListNode head) {
    Set<ListNode> nodesSeen = new HashSet<>();
    while (head != null) {
        if (nodesSeen.contains(head)) {
            return true;
        } else {
            nodesSeen.add(head);
        }
        head = head.next;
    }
    return false;
}
}

什么是HashSet

HashSet实现了Set接口，它不允许集合中有重复的值，当我们提到HashSet时，第一件事情就是在将对象存储在HashSet之前，要先确保对象重写equals()和hashCode()方法，这样才能比较对象的值是否相等，以确保set中没有储存相等的对象。如果我们没有重写这两个方法，将会使用这个方法的默认实现。

Approach 2: Two Pointers

Complexity analysis

Time complexity : O(n)
Space complexity : O(1)

class Solution{
    public boolean hasCycle(ListNode head) {
    if (head == null || head.next == null) {
    //针对链表为null或者说只有一个结点的情况
        return false;
    }
    ListNode slow = head;
    ListNode fast = head.next;
    while (slow != fast) {
    //如果是循环链表，则一个指针快，另一个慢，二者终会相等
    //如果不是循环链表，则到底部之后，返回 false
        if (fast == null || fast.next == null) {
            return false;
        }
        slow = slow.next;
        fast = fast.next.next;
    }
    return true;
}
}