> For the complete documentation index, see [llms.txt](https://imhuay.gitbook.io/studies/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/jian-zhi-offer5302-jian-dan-zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi.md).

# 在排序数组中查找数字

![last modify](https://img.shields.io/static/v1?label=last%20modify\&message=2022-10-14%2014%3A59%3A33\&color=yellowgreen\&style=flat-square) [![](https://img.shields.io/static/v1?label=\&message=%E7%AE%80%E5%8D%95\&color=yellow\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#简单) [![](https://img.shields.io/static/v1?label=\&message=%E5%89%91%E6%8C%87Offer\&color=green\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#剑指offer) [![](https://img.shields.io/static/v1?label=\&message=%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE\&color=blue\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#二分查找)

**问题简述**

```
统计给定数字在排序数组中出现的次数。
```

<details>

<summary><strong>详细描述</strong></summary>

```
统计一个数字在排序数组中出现的次数。

示例 1:
    输入: nums = [5,7,7,8,8,10], target = 8
    输出: 2
示例 2:
    输入: nums = [5,7,7,8,8,10], target = 6
    输出: 0

提示：
    0 <= nums.length <= 10^5
    -10^9 <= nums[i] <= 10^9
    nums 是一个非递减数组
    -10^9 <= target <= 10^9

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi-lcof
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
```

</details>

**思路**

* 二分法分别查找目标值的左右边界；
* 小技巧：如果二分查找的是右边界，那么可以通过查找 `target - 1` 来获得左边界，因为二分查找实际上找的是目标值的插入位置；

<details>

<summary><strong>Python：使用库函数</strong></summary>

```python
import bisect

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        idx_r, idx_l = bisect.bisect_right(nums, target), bisect.bisect_left(nums, target)
        # idx_r, idx_l = bisect.bisect_right(nums, target), bisect.bisect_right(nums, target - 1)
        return idx_r - idx_l
```

</details>

<details>

<summary><strong>Python：不使用库函数</strong></summary>

```python
class Solution:
    def search(self, nums: [int], target: int) -> int:
        
        def bisect(tar):
            l, r = 0, len(nums) - 1
            while l <= r:
                m = (l + r) // 2
                if nums[m] <= tar: 
                    l = m + 1
                else: 
                    r = m - 1
            return l
        
        return bisect(target) - bisect(target - 1)
```

</details>