题目

【判断链表是否有环】

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.

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?

思路

设置一快一慢两个指针，快的一次走两步，慢的一次走一步，若链表中存在环，则两个指针必定会相遇，否则不存在环
时间复杂度 O(n）
空间复杂度 O(1)

代码1

 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if(head == null) return false;
        ListNode fast = head;
        ListNode slow = head;
        while(  fast != null && fast.next != null ){
        	fast = fast.next.next;
        	slow = slow.next;
        	if(fast == slow) return true;

        }
        return false;

    }
}

提交结果

Runtime: 0 ms, faster than 100.00% of Java online submissions for Linked List Cycle.
Memory Usage: 38.7 MB, less than 29.76% of Java online submissions for Linked List Cycle.

代码2

答案的解法，思路一样

public boolean hasCycle(ListNode head) {
    if (head == null || head.next == null) {
        return false;
    }
    ListNode slow = head;
    ListNode fast = head.next;
    while (slow != fast) {
        if (fast == null || fast.next == null) {
            return false;
        }
        slow = slow.next;
        fast = fast.next.next;
    }
    return true;
}

提交结果

Runtime: 0 ms, faster than 100.00% of Java online submissions for Linked List Cycle.
Memory Usage: 38.5 MB, less than 54.54% of Java online submissions for Linked List Cycle.