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LeetCode 141 - Linked List Cycle

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问题描述

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:

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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:

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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:

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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?

Related Topics: Linked List, Two Pointers

原问题: 141. Linked List Cycle

中文翻译版: 141. 环形链表

解决方案

该题可以使用双指针方法进行解决，设定快指针 fast 和慢指针 slow，两指针同时从头节点 head 出发，慢指针每前进一个节点，快指针就前进两个节点，如果链表有环，由于两指针前进速度不同，最终两指针会汇聚在同一个节点，即 fast == slow，否则，快指针会最先到达链表节点，两指针不会汇聚在一起。

参考解题代码 
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#include <iostream>
#include "List.h"
using namespace std;

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool hasCycle(ListNode *head) {
        if (head == NULL)
            return false;

        ListNode *fast, *slow;
        fast = slow = head;
        while (fast != NULL && fast->next != NULL) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast)
                return true;
        }

        return false;
    }
};

int main()
{
    ListNode *node1 = create_list_node(1);
    ListNode *node2 = create_list_node(2);
    ListNode *node3 = create_list_node(3);
    ListNode *node4 = create_list_node(4);
    connect_list_nodes(node1, node2);
    connect_list_nodes(node2, node3);
    connect_list_nodes(node3, node4);
    connect_list_nodes(node4, node2);

    Solution solu;
    cout << solu.hasCycle(node1) << endl;

    return 0;
}