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48 changes: 48 additions & 0 deletions 算法/LeetCode/二分法/69.x的平方根.md
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## [题目](https://leetcode-cn.com/problems/sqrtx/)

实现 `int sqrt(int x)` 函数。

计算并返回 *x* 的平方根,其中 *x* 是非负整数。

由于返回类型是整数,结果只保留整数的部分,小数部分将被舍去。

**示例 1:**

```
输入: 4
输出: 2
```

**示例 2:**

```
输入: 8
输出: 2
说明: 8 的平方根是 2.82842...,
由于返回类型是整数,小数部分将被舍去。
```

## 解法

```java
class Solution {
public int mySqrt(int x) {
long left = 0;
long right = x / 2 + 1;
while(left < right){
long mid = (left + right + 1) >> 1;
long sq = mid * mid;
if(sq > x){
right = mid - 1;
} else {
left = mid;
}
}
return (int)left;
}
}

//牛顿迭代

```

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