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69. Sqrt(x).go
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69. Sqrt(x).go
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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;
// }