> 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/2021/12/jian-zhi-offer4300-kun-nan-1n-zheng-shu-zhong-1-chu-xian-de-ci-shu.md).

# 1～n整数中1出现的次数

![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=%E5%9B%B0%E9%9A%BE\&color=yellow\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/problems/2021/12/pages/R5NyOzkn3qAZy7wCx1pS#困难) [![](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/2021/12/pages/R5NyOzkn3qAZy7wCx1pS#剑指offer) [![](https://img.shields.io/static/v1?label=\&message=%E6%89%BE%E8%A7%84%E5%BE%8B\&color=blue\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/problems/2021/12/pages/R5NyOzkn3qAZy7wCx1pS#找规律)

**问题简述**

> [剑指 Offer 43. 1～n 整数中 1 出现的次数 - 力扣（LeetCode）](https://leetcode-cn.com/problems/1nzheng-shu-zhong-1chu-xian-de-ci-shu-lcof/)

```
输入一个整数 n ，求1～n这n个整数的十进制表示中1出现的次数。
```

<details>

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

```
输入一个整数 n ，求1～n这n个整数的十进制表示中1出现的次数。

例如，输入12，1～12这些整数中包含1 的数字有1、10、11和12，1一共出现了5次。

示例 1：
    输入：n = 12
    输出：5
示例 2：
    输入：n = 13
    输出：6

限制：
    1 <= n < 2^31

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/1nzheng-shu-zhong-1chu-xian-de-ci-shu-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**

* 找规律的题目，很费时，建议直接看：[1～n 整数中 1 出现的次数（清晰图解） - Krahets](https://leetcode-cn.com/problems/1nzheng-shu-zhong-1chu-xian-de-ci-shu-lcof/solution/mian-shi-ti-43-1n-zheng-shu-zhong-1-chu-xian-de-2/)

<details>

<summary><strong>Python</strong></summary>

```python
class Solution:
    def countDigitOne(self, n: int) -> int:
        # 初始化一些变量
        digit, ret = 1, 0
        hi, cur, lo = n // 10, n % 10, 0

        while hi != 0 or cur != 0:
            if cur == 0:
                ret += hi * digit
            elif cur == 1:
                ret += hi * digit + lo + 1
            else:
                ret += (hi + 1) * digit
            lo += cur * digit
            cur = hi % 10
            hi //= 10
            digit *= 10
        return ret
```

</details>