LeetCode Top 100 Liked Questions 141. Linked List Cycle (Java版; Easy)

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LeetCode Top 100 Liked Questions 141. Linked List Cycle (Java版; Easy)

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.

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.

第一次做; 快慢指针; 理解初始化和循环终止条件; 快指针每次移动两步, 判断条件是固定的if(fast==null || fast.next==null), 找链表中点的时候也用到了快慢指针

• 时间复杂度O(N)
• 空间复杂度O(1)

/*
数学证明?
题解中的定性分析:
快跑者最终一定会追上慢跑者。这是为什么呢？
考虑下面这种情况（记作情况 A）,假如快跑者只落后慢跑者一步，在下一次迭代中，它们就会分别跑了一步或两步并相遇。
其他情况又会怎样呢？例如，我们没有考虑快跑者在慢跑者之后两步或三步的情况。但其实不难想到，
因为在下一次或者下下次迭代后，又会变成上面提到的情况 A。
*/
public class Solution {
    public boolean hasCycle(ListNode head) {
        //input check
        if(head==null || head.next==null)
            return false;
        //
        ListNode slow = head, fast = head.next;
        while(slow != fast){
            if(fast==null || fast.next==null)
                return false;
            slow = slow.next;
            fast = fast.next.next;
        }
        return true;
    }
}