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144 lines
3.9 KiB
Markdown
144 lines
3.9 KiB
Markdown
> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
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>
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> 感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。
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>
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> 另外希望手机阅读的同学可以来我的 <u>[**公众号：程序厨**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u> 两个平台同步，想要和题友一起刷题，互相监督的同学，可以在我的小屋点击<u>[**刷题小队**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u>进入。
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### [141. 环形链表](https://leetcode-cn.com/problems/linked-list-cycle/)
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下面我们再来了解一种双指针，我们称之为快慢指针，顾名思义一个指针速度快，一个指针速度慢。
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#### 题目描述
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> 给定一个链表，判断链表中是否有环。pos 代表环的入口，若为-1，则代表无环。
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>
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> 如果链表中存在环，则返回 true 。否则，返回 false 。
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示例 1：
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> 输入：head = [3,2,0,-4], pos = 1
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> 输出：true
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> 解释：链表中有一个环，其尾部连接到第二个节点。
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#### 题目解析
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题目很容易理解，让我们判断链表中是否有环，我们只需通过我们的快慢指针即可，我们试想一下，如果链表中有环的话，一个速度快的指针和一个速度慢的指针在环中运动的话，若干圈后快指针肯定可以追上慢指针的。这是一定的。
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好啦，做题思路已经有了，让我们一起看一下代码的执行过程吧。
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**动画模拟**
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**题目代码**
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Java Code:
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```java
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public class Solution {
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public boolean hasCycle(ListNode head) {
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ListNode fast = head;
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ListNode low = head;
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while (fast != null && fast.next != null) {
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fast = fast.next.next;
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low = low.next;
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if (fast == low) {
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return true;
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}
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}
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return false;
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}
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}
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```
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C++ Code:
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```cpp
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class Solution {
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public:
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bool hasCycle(ListNode *head) {
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ListNode * fast = head;
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ListNode * slow = head;
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while (fast != nullptr && fast->next != nullptr) {
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fast = fast->next->next;
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slow = slow->next;
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if (fast == slow) {
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return true;
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}
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}
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return false;
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}
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};
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```
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JS Code:
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```javascript
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var hasCycle = function (head) {
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let fast = head;
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let slow = head;
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while (fast && fast.next) {
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fast = fast.next.next;
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slow = slow.next;
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if (fast === slow) {
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return true;
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}
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}
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return false;
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};
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```
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Python Code:
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```python
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class Solution:
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def hasCycle(self, head: ListNode) -> bool:
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fast = head
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slow = head
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while fast and fast.next:
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fast = fast.next.next
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slow = slow.next
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if fast == slow:
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return True
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return False
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```
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Swift Code：
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```swift
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class Solution {
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func hasCycle(_ head: ListNode?) -> Bool {
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var fast = head, slow = head
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while fast != nil && fast?.next != nil {
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fast = fast?.next?.next
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slow = slow?.next
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if fast === slow {
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return true
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}
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}
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return false
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}
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}
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```
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Go Code:
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```go
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func hasCycle(head *ListNode) bool {
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if head == nil { return false }
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s, f := head, head
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for f != nil && f.Next != nil {
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s = s.Next
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f = f.Next.Next
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if s == f {
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return true
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}
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}
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return false
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}
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```
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