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  • leetcode 141 Linked List Cycle Hash fast and slow pointer

    Problem describe:https://leetcode.com/problems/linked-list-cycle/

    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?


    Ac Code: (Hash)

     1 /**
     2  * Definition for singly-linked list.
     3  * struct ListNode {
     4  *     int val;
     5  *     ListNode *next;
     6  *     ListNode(int x) : val(x), next(NULL) {}
     7  * };
     8  */
     9 class Solution {
    10 public:
    11     bool hasCycle(ListNode *head) {
    12         unordered_set<ListNode*> visit;
    13         while(head)
    14         {
    15             if(visit.count(head)!=0) return true;
    16             visit.insert(head);
    17             head = head->next;
    18         }
    19         return false;
    20     }
    21 };

     Fast and Slow Pointer

     1 /**
     2  * Definition for singly-linked list.
     3  * struct ListNode {
     4  *     int val;
     5  *     ListNode *next;
     6  *     ListNode(int x) : val(x), next(NULL) {}
     7  * };
     8  */
     9 class Solution {
    10 public:
    11     bool hasCycle(ListNode *head) {
    12         ListNode *slow = head;
    13         ListNode *fast = head;
    14         while(fast){
    15             if(!fast->next) return false;
    16             fast = fast->next->next;
    17             slow = slow->next;
    18             if(slow == fast) return true;
    19         }
    20         return false;
    21     }
    22 };
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  • 原文地址:https://www.cnblogs.com/CS-WLJ/p/11181770.html
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