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pull merged 1 commit into AlgorithmAndLeetCode:main from itcharge:main
Jul 11, 2022
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29 changes: 24 additions & 5 deletions Solutions/0374. 猜数字大小.md
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## 题目大意

猜数字游戏。给定一个整数 n 和一个接口 `def guess(num: int) -> int:`,题目会从 1~n 中随机选取一个数 x。我们只能通过调用接口来判断自己猜测的数是否正确。要求返回题目选取的数字 x。
**描述**:猜数字游戏。给定一个整数 `n` 和一个接口 `def guess(num: int) -> int:`,题目会从 `1` ~ `n` 中随机选取一个数 `x`。我们只能通过调用接口来判断自己猜测的数是否正确。

**要求**:要求返回题目选取的数字 `x`。

**说明**:

- `def guess(num: int) -> int:` 返回值:
- $-1$:我选出的数字比你猜的数字小,即 $pick < num$;
- 1ドル$:我选出的数字比你猜的数字大 $pick > num$;
- 0ドル$:我选出的数字和你猜的数字一样。恭喜!你猜对了!$pick == num$。


**示例**:

```Python
输入 n = 10, pick = 6
输出 6
```

## 解题思路

利用两个指针 left、right。left 指向数字 1,right 指向数组 n。每次从中间开始调用接口猜测是否正确。
### 思路 1:二分查找

利用两个指针 `left`、`right`。`left` 指向数字 `1`,`right` 指向数字 `n`。每次从中间开始调用接口猜测是否正确。

- 如果猜测的数比选中的数大,则将 right 向左移,继续从中间调用接口猜测;
- 如果猜测的数比选中的数小,则将 left 向右移,继续从中间调用的接口猜测;
- 如果猜测的数比选中的数大,则将 `right` 向左移,令 `right = mid - 1`,继续从中间调用接口猜测;
- 如果猜测的数比选中的数小,则将 `left` 向右移,令 `left = mid + 1`,继续从中间调用的接口猜测;
- 如果猜测正确,则直接返回该数。

## 代码
### 思路 1:二分查找代码

```Python
class Solution:
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