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Solution141.hpp
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67 lines (55 loc) · 1.94 KB
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//
// Solution141.hpp
// Algorithm
//
// Created by Pancf on 2019/12/21.
// Copyright © 2019 Pancf. All rights reserved.
//
#ifndef Solution141_hpp
#define Solution141_hpp
#include <stdio.h>
class Solution141 {
/**
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?
================================================================================================
Accept details:
Runtime: 8 ms, faster than 97.17% of C++ online submissions for Linked List Cycle.
Memory Usage: 9.6 MB, less than 100.00% of C++ online submissions for Linked List Cycle.
思路：用快慢指针即可，慢指针每个loop向前走一步，快指针向前走两步，若有环则快指针一定会追上慢指针，若无环则快指针一定先到达链表尾部
*/
public:
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
bool hasCycle(ListNode *head);
static void test() {
ListNode node1{1};
ListNode node2{2};
ListNode node3{3};
ListNode node4{4};
node1.next = &node2;
node2.next = &node3;
node3.next = &node4;
node4.next = &node2;
Solution141 s = Solution141();
printf("%d", s.hasCycle(&node1));
}
};
#endif /* Solution141_hpp */