如何实现sqrt()
2024-09-06 05:15:36
jdk中实现sqrt()是native方法，没法看到具体的实现细节，所以自己整理下，以便后续查阅。
1、暴力法，从0开始每次增加1e-6，直到非常接近
2、牛顿法，求n的平方根
while(abs(x-x_pre)>1e-6){
x_pre = x;
x = (x+n/x)/2;
}
return x;
3、二分法
4、快速平方根倒数，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 >> ); // 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;
}
java版本
public static float invSqrt(float x) {
float xhalf = 0.5f*x;
int i = Float.floatToIntBits(x);
i = 0x5f3759df - (i>>);
x = Float.intBitsToFloat(i);
x = x*(1.5f - xhalf*x*x);
return x;
}
5、快速计算(int)(sqrt(x))，利用空间换时间
public class APIsqrt2 {
final static int[] table = { 0, 16, 22, 27, 32, 35, 39, 42, 45, 48, 50, 53,
55, 57, 59, 61, 64, 65, 67, 69, 71, 73, 75, 76, 78, 80, 81, 83, 84,
86, 87, 89, 90, 91, 93, 94, 96, 97, 98, 99, 101, 102, 103, 104,
106, 107, 108, 109, 110, 112, 113, 114, 115, 116, 117, 118, 119,
120, 121, 122, 123, 124, 125, 126, 128, 128, 129, 130, 131, 132,
133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 144,
145, 146, 147, 148, 149, 150, 150, 151, 152, 153, 154, 155, 155,
156, 157, 158, 159, 160, 160, 161, 162, 163, 163, 164, 165, 166,
167, 167, 168, 169, 170, 170, 171, 172, 173, 173, 174, 175, 176,
176, 177, 178, 178, 179, 180, 181, 181, 182, 183, 183, 184, 185,
185, 186, 187, 187, 188, 189, 189, 190, 191, 192, 192, 193, 193,
194, 195, 195, 196, 197, 197, 198, 199, 199, 200, 201, 201, 202,
203, 203, 204, 204, 205, 206, 206, 207, 208, 208, 209, 209, 210,
211, 211, 212, 212, 213, 214, 214, 215, 215, 216, 217, 217, 218,
218, 219, 219, 220, 221, 221, 222, 222, 223, 224, 224, 225, 225,
226, 226, 227, 227, 228, 229, 229, 230, 230, 231, 231, 232, 232,
233, 234, 234, 235, 235, 236, 236, 237, 237, 238, 238, 239, 240,
240, 241, 241, 242, 242, 243, 243, 244, 244, 245, 245, 246, 246,
247, 247, 248, 248, 249, 249, 250, 250, 251, 251, 252, 252, 253,
253, 254, 254, 255 };
/**
* A faster replacement for (int)(java.lang.Math.sqrt(x)). Completely
* accurate for x < 2147483648 (i.e. 2^31)...
*/
static int sqrt(int x) {
int xn;
if (x >= 0x10000) {
if (x >= 0x1000000) {
if (x >= 0x10000000) {
if (x >= 0x40000000) {
xn = table[x >> 24] << 8;
} else {
xn = table[x >> 22] << 7;
}
} else {
if (x >= 0x4000000) {
xn = table[x >> 20] << 6;
} else {
xn = table[x >> 18] << 5;
}
}
xn = (xn + 1 + (x / xn)) >> 1;
xn = (xn + 1 + (x / xn)) >> 1;
return ((xn * xn) > x) ? --xn : xn;
} else {
if (x >= 0x100000) {
if (x >= 0x400000) {
xn = table[x >> 16] << 4;
} else {
xn = table[x >> 14] << 3;
}
} else {
if (x >= 0x40000) {
xn = table[x >> 12] << 2;
} else {
xn = table[x >> 10] << 1;
}
}
xn = (xn + 1 + (x / xn)) >> 1;
return ((xn * xn) > x) ? --xn : xn;
}
} else {
if (x >= 0x100) {
if (x >= 0x1000) {
if (x >= 0x4000) {
xn = (table[x >> 8]) + 1;
} else {
xn = (table[x >> 6] >> 1) + 1;
}
} else {
if (x >= 0x400) {
xn = (table[x >> 4] >> 2) + 1;
} else {
xn = (table[x >> 2] >> 3) + 1;
}
}
return ((xn * xn) > x) ? --xn : xn;
} else {
if (x >= 0) {
return table[x] >> 4;
}
}
}
return -1;
}
public static void main(String[] args){
System.out.println(sqrt(65));
}
}
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