0069. Sqrtx

# 69. Sqrt(x)#

## 题目 #

Implement `int sqrt(int x)`.

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

``````Input: 4
Output: 2
``````

Example 2:

``````Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since
the decimal part is truncated, 2 is returned.
``````

## 解题思路 #

• 题目要求求出根号 x

• 根据题意，根号 x 的取值范围一定在 `[0,x]` 之间，这个区间内的值是递增有序的，有边界的，可以用下标访问的，满足这三点正好也就满足了二分搜索的 3 大条件。所以解题思路一，二分搜索。

• 解题思路二，牛顿迭代法。求根号 x，即求满足 `x^2 - n = 0` 方程的所有解。

## 代码 #

``````
package leetcode

// 解法一 二分
func mySqrt(x int) int {
if x == 0 {
return 0
}
left, right, res := 1, x, 0
for left <= right {
mid := left + ((right - left) >> 1)
if mid < x/mid {
left = mid + 1
res = mid
} else if mid == x/mid {
return mid
} else {
right = mid - 1
}
}
return res
}

// 解法二 牛顿迭代法 https://en.wikipedia.org/wiki/Integer_square_root
func mySqrt1(x int) int {
r := x
for r*r > x {
r = (r + x/r) / 2
}
return r
}

// 解法三 Quake III 游戏引擎中有一种比 STL 的 sqrt 快 4 倍的实现 https://en.wikipedia.org/wiki/Fast_inverse_square_root
// float Q_rsqrt( float number )
// {
// 	long i;
// 	float x2, y;
// 	const float threehalfs = 1.5F;

// 	x2 = number * 0.5F;
// 	y  = number;
// 	i  = * ( long * ) &y;                       // evil floating point bit level hacking
// 	i  = 0x5f3759df - ( i >> 1 );               // what the fuck?
// 	y  = * ( float * ) &i;
// 	y  = y * ( threehalfs - ( x2 * y * y ) );   // 1st iteration
// //	y  = y * ( threehalfs - ( x2 * y * y ) );   // 2nd iteration, this can be removed
// 	return y;
// }

``````

Sep 6, 2020