环形链表

Discussion is really funny！

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?

简单解释一下

这题目综述写得糟糕的很，很多人（包括我）看不懂到底是要干什么。其实就是给你一个链表，让你判断是不是环形链表，环形链表永远跑不到尾巴。

Mine

/**
 * 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) {
        ListNode slow = head;
        if(head == null ) return false;
        ListNode fast = head.next;
        if (fast == null) {
            return false;
        }
        try{
          do {
            slow = slow.next;
            fast = fast.next.next;
            if (fast == slow) {
                return true;
          			}   
        	 } while (fast != null);
   
        }catch(Exception e){
            return false;
        }
        return false;
    }
}

双指针法注意.next.next找不到的情况。

Discussion

今天才看到有讨论区，真是人才辈出！

Delay

bool hasCycle(struct ListNode *head) {
    if(head){
        int times = 0;
        struct ListNode *root = head;
        while(root->next){
            root = root->next;
            times++;
            if(times>10000){
                return true;
            }
        }
    }
    return false;

" 我帮你们测试好了，样例最多的有8029个数据 "

public class Solution {
    public boolean hasCycle(ListNode head) {
        int count=8029;
        
        while(head!=null&&count>0){
            head=head.next;
            count--;
        }
        if(head==null)
            return false;
            
        return true;
        
        
    }
}

”想要取巧,bjfuvth什么的怪字符串不可取“

var hasCycle = function(head) {
    let s = Symbol('fuck')
    while(head){
        if(head[s]) return true
        else head[s] = true
        head = head.next
    }
    return false
};

???

var hasCycle = function(head) {
    try{
        JSON.stringify(head);
    }
    catch(e){
        return true;
    }
    return false;
};