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C++實(shí)現(xiàn)LeetCode(141.單鏈表中的環(huán))

 更新時(shí)間:2021年07月22日 16:23:22   作者:Grandyang  
這篇文章主要介紹了C++實(shí)現(xiàn)LeetCode(141.單鏈表中的環(huán)),本篇文章通過簡要的案例,講解了該項(xiàng)技術(shù)的了解與使用,以下就是詳細(xì)內(nèi)容,需要的朋友可以參考下

[LeetCode] 141. Linked List Cycle 單鏈表中的環(huán)

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?

這道題是快慢指針的經(jīng)典應(yīng)用。只需要設(shè)兩個(gè)指針,一個(gè)每次走一步的慢指針和一個(gè)每次走兩步的快指針,如果鏈表里有環(huán)的話,兩個(gè)指針最終肯定會(huì)相遇。實(shí)在是太巧妙了,要是我肯定想不出來。代碼如下:

C++ 解法:

class Solution {
public:
    bool hasCycle(ListNode *head) {
        ListNode *slow = head, *fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) return true;
        }
        return false;
    }
};

Java 解法:

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

Github 同步地址:

https://github.com/grandyang/leetcode/issues/141

類似題目:

Linked List Cycle II

Happy Number

參考資料:

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

https://leetcode.com/problems/linked-list-cycle/discuss/44489/O(1)-Space-Solution

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