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  • [Java]LeetCode141. 环形链表 | Linked List Cycle

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

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


    给定一个链表,判断链表中是否有环。

    进阶:
    你能否不使用额外空间解决此题?


     0ms 
     1 /**
     2  * Definition for singly-linked list.
     3  * class ListNode {
     4  *     int val;
     5  *     ListNode next;
     6  *     ListNode(int x) {
     7  *         val = x;
     8  *         next = null;
     9  *     }
    10  * }
    11  */
    12 public class Solution {
    13     public boolean hasCycle(ListNode head) {
    14         if(head == null) return false;
    15         ListNode fast = head;
    16         ListNode slow = head;
    17         do{
    18             if(fast.next == null || fast.next.next == null) return false;
    19             fast = fast.next.next;
    20             slow = slow.next;
    21         }while(fast != slow);
    22         return true;
    23     }
    24 }

    0ms

     1 /**
     2  * Definition for singly-linked list.
     3  * class ListNode {
     4  *     int val;
     5  *     ListNode next;
     6  *     ListNode(int x) {
     7  *         val = x;
     8  *         next = null;
     9  *     }
    10  * }
    11  */
    12 public class Solution {
    13     public boolean hasCycle(ListNode head) {
    14     if(head==null) return false;
    15     ListNode walker = head;
    16     ListNode runner = head;
    17     while(runner.next!=null && runner.next.next!=null) {
    18         walker = walker.next;
    19         runner = runner.next.next;
    20         if(walker==runner) return true;
    21     }
    22     return false;
    23 }
    24 }

    1ms

     1 /**
     2  * Definition for singly-linked list.
     3  * class ListNode {
     4  *     int val;
     5  *     ListNode next;
     6  *     ListNode(int x) {
     7  *         val = x;
     8  *         next = null;
     9  *     }
    10  * }
    11  */
    12 public class Solution {
    13 public static boolean hasCycle(ListNode head) {
    14         if (head == null || head.next == null) {
    15             return false;
    16         }
    17         ListNode slow = head;
    18         ListNode fast = head.next;
    19         int f = 1;//上面两行赋值已前进了一次
    20         while (slow != fast) {
    21             if (fast == null || fast.next == null) {
    22                 return false;
    23             }
    24             slow = slow.next;
    25             fast = fast.next.next;
    26             f++;
    27         }
    28         System.out.println("LinkedList.size:"+f);
    29         return true;
    30     }
    31 }

    4ms

     1 public class Solution {
     2     public boolean hasCycle(ListNode head) {
     3         HashSet<ListNode> visited = new HashSet<>();
     4         
     5         ListNode curr = head;
     6         while (curr != null) {
     7             if (visited.contains(curr))
     8                 return true;
     9             visited.add(curr);
    10             curr = curr.next;
    11         }
    12         
    13         return false;
    14     }
    15 }

    5ms

     1 /**
     2  * Definition for singly-linked list.
     3  * class ListNode {
     4  *     int val;
     5  *     ListNode next;
     6  *     ListNode(int x) {
     7  *         val = x;
     8  *         next = null;
     9  *     }
    10  * }
    11  */
    12 public class Solution {
    13     public boolean hasCycle(ListNode head) {
    14         Set<ListNode> set = new HashSet<>();
    15         while(head != null){
    16             if(set.contains(head)){
    17                 return true;
    18             } else{
    19                 set.add(head);
    20             }
    21             head = head.next;
    22         }
    23         return false;
    24     }
    25 }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/9714958.html
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