[LeetCode]92. Linked List Cycle判断链表是否有环

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

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

 

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解法1：使用两个指针p1, p2。p1从表头开始一步一步往前走，遇到null则说明没有环，返回false；p1每走一步，p2从头开始走，如果遇到p2==p1.next，则说明有环true，如果遇到p2==p1，则说明暂时没有环，继续循环。但是这个时间复杂度O(n^2)，会超时Time Limit Exceeded

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

 

解法2：使用快慢指针，慢指针每次前移一个节点，快指针每次前移两个节点，如果后面两个指针相交，就说明存在环。时间复杂度O(n)。

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