題目簡介

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

Follow up:
Can you solve it without using extra space?

自己的解法

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

Hot解法

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

runner.next!=null && runner.next.next!=null

• 我們的解法更應該追求一致性，儘量都可以放到迴圈裡處理，而少一些需要單獨拿出來處理的操作。