solution.md

# 在排序数组中查找元素的第一个和最后一个位置

<p>给定一个按照升序排列的整数数组 <code>nums</code>，和一个目标值 <code>target</code>。找出给定目标值在数组中的开始位置和结束位置。</p>
<p>如果数组中不存在目标值 <code>target</code>，返回 <code>[-1, -1]</code>。</p>
<p><strong>进阶：</strong></p>
<ul>
    <li>你可以设计并实现时间复杂度为 <code>O(log n)</code> 的算法解决此问题吗？</li>
</ul>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre><strong>输入：</strong>nums = [5,7,7,8,8,10], target = 8<strong><br />输出：</strong>[3,4]</pre>
<p><strong>示例 2：</strong></p>
<pre><strong>输入：</strong>nums = [5,7,7,8,8,10], target = 6<strong><br />输出：</strong>[-1,-1]</pre>
<p><strong>示例 3：</strong></p>
<pre><strong>输入：</strong>nums = [], target = 0<strong><br />输出：</strong>[-1,-1]</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
    <li><code>0 <= nums.length <= 10<sup>5</sup></code></li>
    <li><code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code></li>
    <li><code>nums</code> 是一个非递减数组</li>
    <li><code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code></li>
</ul>

## template

```cpp
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
	vector<int> searchRange(vector<int> &nums, int target)
	{
		vector<int> res;
		res.push_back(binary_search_begin(nums, target));
		res.push_back(binary_search_end(nums, target));
		return res;
	}
private:
	int binary_search_begin(vector<int> nums, int target)
	{
		int lo = -1;
		int hi = nums.size();
		while (lo + 1 < hi)
		{
			int mid = lo + (hi - lo) / 2;
			if (target > nums[mid])
			{
				lo = mid;
			}
			else
			{
				hi = mid;
			}
		}
		if (hi == nums.size() || nums[hi] != target)
		{
			return -1;
		}
		else
		{
			return hi;
		}
	}
	int binary_search_end(vector<int> nums, int target)
	{
		int lo = -1;
		int hi = nums.size();
		while (lo + 1 < hi)
		{
			int mid = lo + (hi - lo) / 2;
			if (target < nums[mid])
			{
				hi = mid;
			}
			else
			{
				lo = mid;
			}
		}
		if (lo == -1 || nums[lo] != target)
		{
			return -1;
		}
		else
		{
			return lo;
		}
	}
};
```

## 答案

```cpp

```

## 选项

### A

```cpp

```

### B

```cpp

```

### C

```cpp

```