# 环形链表 II

<p>给定一个链表，返回链表开始入环的第一个节点。 如果链表无环，则返回 <code>null</code>。</p>

<p>为了表示给定链表中的环，我们使用整数 <code>pos</code> 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 <code>pos</code> 是 <code>-1</code>，则在该链表中没有环。<strong>注意，<code>pos</code> 仅仅是用于标识环的情况，并不会作为参数传递到函数中。</strong></p>

<p><strong>说明：</strong>不允许修改给定的链表。</p>

<p><strong>进阶：</strong></p>

<ul>
	<li>你是否可以使用 <code>O(1)</code> 空间解决此题？</li>
</ul>



<p><strong>示例 1：</strong></p>

<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/07/circularlinkedlist.png" style="height: 97px; width: 300px;" /></p>

<pre>
<strong>输入：</strong>head = [3,2,0,-4], pos = 1

<strong>输出：</strong>返回索引为 1 的链表节点

<strong>解释：</strong>链表中有一个环，其尾部连接到第二个节点。
</pre>

<p><strong>示例 2：</strong></p>

<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/07/circularlinkedlist_test2.png" style="height: 74px; width: 141px;" /></p>

<pre>
<strong>输入：</strong>head = [1,2], pos = 0

<strong>输出：</strong>返回索引为 0 的链表节点

<strong>解释：</strong>链表中有一个环，其尾部连接到第一个节点。
</pre>

<p><strong>示例 3：</strong></p>

<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/07/circularlinkedlist_test3.png" style="height: 45px; width: 45px;" /></p>

<pre>
<strong>输入：</strong>head = [1], pos = -1

<strong>输出：</strong>返回 null

<strong>解释：</strong>链表中没有环。
</pre>



<p><strong>提示：</strong></p>

<ul>
	<li>链表中节点的数目范围在范围 <code>[0, 10<sup>4</sup>]</code> 内</li>
	<li><code>-10<sup>5</sup> <= Node.val <= 10<sup>5</sup></code></li>
	<li><code>pos</code> 的值为 <code>-1</code> 或者链表中的一个有效索引</li>
</ul>

<p>以下<span style="color:red">错误</span>的选项是？</p>

## aop

### before

```c
#include <bits/stdc++.h>
using namespace std;

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};
```

### after

```c

```

## 答案

```c
class Solution
{
public:
    vector<int> searchRange(vector<int> &nums, int target)
    {
        int left = 0;
        int right = nums.size() - 1;
        vector<int> res = {-1, -1};
        bool find = false;
        while (right >= left && find == false && nums[left] <= target)
        {
            int mid = (left + right) / 2;
            if (nums[mid] > target)
                right = mid - 1;
            else if (nums[mid] < target)
                left = mid + 1;
            else
            {
                find = true;
                while (mid > 0 && nums[mid - 1] == target)
                    mid++;
                res[0] = mid;
                while (mid < nums.size() - 1 && nums[mid + 1] == target)
                    mid--;
                res[1] = mid;
            }
        }
        return res;
    }
};
```
## 选项


### A

```c
class Solution
{
public:
    ListNode *detectCycle(ListNode *head)
    {
        set<ListNode *> node_set;
        while (head)
        {
            if (node_set.find(head) != node_set.end())
            {
                return head;
            }
            node_set.insert(head);
            head = head->next;
        }
        return NULL;
    }
};
```

### B

```c
class Solution
{
public:
    ListNode *detectCycle(ListNode *head)
    {
        ListNode *slow = head;
        ListNode *fast = head;
        ListNode *meet = NULL;
        while (slow && fast && fast->next)
        {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast)
            {
                meet = slow;
                break;
            }
        }
        while (head && meet)
        {
            if (meet == head)
            {
                break;
            }
            head = head->next;
            meet = meet->next;
        }
        return meet;
    }
};
```

### C

```c
class Solution
{
public:
    ListNode *detectCycle(ListNode *head)
    {
        if (!head || !head->next)
            return NULL;
        ListNode *slow = head, *fast = head;
        while (fast && fast->next)
        {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast)
                break;
        }
        if (slow != fast)
            return NULL;
        slow = head;
        while (slow != fast)
        {
            slow = slow->next;
            fast = fast->next;
        }
        return slow;
    }
};
```
