LeetCode解题报告（285）-- 141. Linked List Cycle

Problem

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

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to. Note that pos is not passed as a parameter.

Return true if there is a cycle in the linked list. Otherwise, return false.

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 the tail connects to the 1st node (0-indexed).

Example 2:

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Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.

Example 3:

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Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

Constraints:

The number of the nodes in the list is in the range [0, 104].
-105 <= Node.val <= 105
pos is -1 or a valid index in the linked-list.

Follow up: Can you solve it using O(1) (i.e. constant) memory?

Analysis

题目给出一个链表，要求判断是否存在环。链表中是否存在环这个问题一般都睡通过快慢指针来解决的，慢指针一次走一步，快指针一次走两步，如果两个指针有相遇，说明存在环。

Solution

无。

Code

/**
 * 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) {
        ListNode* p1 = head, *p2 = head;
        while (p2 && p2->next) {
            p1 = p1->next;
            p2 = p2->next->next;
            if (p1 == p2) {
                return true;
            }
        }
        return false;
    }
};

Summary

在链表中使用快慢指针是非常常见的解题思路，也是后续解决许多同类型题目的基础。这道题这道题目的分享到这里，谢谢您的支持！