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## [题目](https://leetcode-cn.com/problems/sqrtx/) | ||
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实现 `int sqrt(int x)` 函数。 | ||
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计算并返回 *x* 的平方根,其中 *x* 是非负整数。 | ||
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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。 | ||
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**示例 1:** | ||
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``` | ||
输入: 4 | ||
输出: 2 | ||
``` | ||
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**示例 2:** | ||
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``` | ||
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。 | ||
``` | ||
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## 解法 | ||
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```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; | ||
} | ||
} | ||
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//牛顿迭代 | ||
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``` | ||
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