【LeetCode】141. Linked List Cycle 解题报告（Java & Python & C++）
作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/
[LeetCode]
题目地址：https://leetcode.com/problems/linked-list-cycle/
Total Accepted: 102417 Total Submissions: 277130 Difficulty: Easy
题目描述
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:
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?
题目大意
判断单链表里是否有环。
解题方法
双指针
双指针的方法。
思路是两个指针，一个每次走两步，一个每次走一步，循环下去，只要两者能够重逢说明有环。
Java代码如下：
public class Solution {
public boolean hasCycle(ListNode head) {
if(head==null) return false;
ListNode fast = head;
ListNode slow = head;
while(slow!=null){
if(fast.next==null || fast.next.next==null) return false;
fast=fast.next.next;
slow=slow.next;
if(fast==slow) break;
}
return true;
}
}
AC:1ms
看了官方解答之后，发现可以优化，优化如下：
public class Solution {
public boolean hasCycle(ListNode head) {
if(head==null||head.next==null) return false;
ListNode fast = head.next;
ListNode slow = head;
while(slow!=fast){
if(fast.next==null || fast.next.next==null) return false;
fast=fast.next.next;
slow=slow.next;
}
return true;
}
}
我想的是只要走的慢的这个不为空的话，就一直走好了。 官方解答想的是两者不重合就一直走。
二刷，Python。
上python版本的。
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def hasCycle(self, head):
"""
:type head: ListNode
:rtype: bool
"""
slow, fast = head, head
while fast and fast.next:
fast = fast.next.next
slow = slow.next
if slow == fast:
return True
return False
三刷，python.
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def hasCycle(self, head):
"""
:type head: ListNode
:rtype: bool
"""
if not head: return False
slow, fast = head, head.next
while fast and fast.next:
if fast == slow:
return True
fast = fast.next.next
slow = slow.next
return False
四刷，C++。注意C++里面全部用的是指针操作。代码如下：
/**
* 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) return false;
ListNode* fast = head;
ListNode* slow = head;
while (fast && fast->next) {
fast = fast->next->next;
slow = slow->next;
if (fast == slow)
return true;
}
return false;
}
};
保存已经走过的路径
官方解答的HashTable的方法。记录下来哪些已经走了，只要走到之前走过的节点说明有环。
public class Solution {
public boolean hasCycle(ListNode head) {
HashSet<ListNode> hash=new HashSet();
while(head!=null){
if(hash.contains(head)){
return true;
}else{
hash.add(head);
}
head=head.next;
}
return false;
}
}
AC:10ms
日期
2016/5/2 17:24:37
2018 年 11 月 24 日 —— 周六快乐
2019 年 1 月 11 日 —— 小光棍节？
