69. Sqrt(x).go

package leetcode

// 解法一 二分, 找到最后一个满足 n^2 <= x 的整数n
func mySqrt(x int) int {
	l, r := 0, x
	for l < r {
		mid := (l + r + 1) / 2
		if mid*mid > x {
			r = mid - 1
		} else {
			l = mid
		}
	}
	return l
}

// 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;
// }