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 >> 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; }
java版本
public static float invSqrt(float x) { float xhalf = 0.5f*x; int i = Float.floatToIntBits(x); i = 0x5f3759df - (i>>1); 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
1 public class APIsqrt2 { 2 final static int[] table = { 0, 16, 22, 27, 32, 35, 39, 42, 45, 48, 50, 53, 3 55, 57, 59, 61, 64, 65, 67, 69, 71, 73, 75, 76, 78, 80, 81, 83, 84, 4 86, 87, 89, 90, 91, 93, 94, 96, 97, 98, 99, 101, 102, 103, 104, 5 106, 107, 108, 109, 110, 112, 113, 114, 115, 116, 117, 118, 119, 6 120, 121, 122, 123, 124, 125, 126, 128, 128, 129, 130, 131, 132, 7 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 144, 8 145, 146, 147, 148, 149, 150, 150, 151, 152, 153, 154, 155, 155, 9 156, 157, 158, 159, 160, 160, 161, 162, 163, 163, 164, 165, 166, 10 167, 167, 168, 169, 170, 170, 171, 172, 173, 173, 174, 175, 176, 11 176, 177, 178, 178, 179, 180, 181, 181, 182, 183, 183, 184, 185, 12 185, 186, 187, 187, 188, 189, 189, 190, 191, 192, 192, 193, 193, 13 194, 195, 195, 196, 197, 197, 198, 199, 199, 200, 201, 201, 202, 14 203, 203, 204, 204, 205, 206, 206, 207, 208, 208, 209, 209, 210, 15 211, 211, 212, 212, 213, 214, 214, 215, 215, 216, 217, 217, 218, 16 218, 219, 219, 220, 221, 221, 222, 222, 223, 224, 224, 225, 225, 17 226, 226, 227, 227, 228, 229, 229, 230, 230, 231, 231, 232, 232, 18 233, 234, 234, 235, 235, 236, 236, 237, 237, 238, 238, 239, 240, 19 240, 241, 241, 242, 242, 243, 243, 244, 244, 245, 245, 246, 246, 20 247, 247, 248, 248, 249, 249, 250, 250, 251, 251, 252, 252, 253, 21 253, 254, 254, 255 }; 22 23 /** 24 * A faster replacement for (int)(java.lang.Math.sqrt(x)). Completely 25 * accurate for x < 2147483648 (i.e. 2^31)... 26 */ 27 static int sqrt(int x) { 28 int xn; 29 30 if (x >= 0x10000) { 31 if (x >= 0x1000000) { 32 if (x >= 0x10000000) { 33 if (x >= 0x40000000) { 34 xn = table[x >> 24] << 8; 35 } else { 36 xn = table[x >> 22] << 7; 37 } 38 } else { 39 if (x >= 0x4000000) { 40 xn = table[x >> 20] << 6; 41 } else { 42 xn = table[x >> 18] << 5; 43 } 44 } 45 46 xn = (xn + 1 + (x / xn)) >> 1; 47 xn = (xn + 1 + (x / xn)) >> 1; 48 return ((xn * xn) > x) ? --xn : xn; 49 } else { 50 if (x >= 0x100000) { 51 if (x >= 0x400000) { 52 xn = table[x >> 16] << 4; 53 } else { 54 xn = table[x >> 14] << 3; 55 } 56 } else { 57 if (x >= 0x40000) { 58 xn = table[x >> 12] << 2; 59 } else { 60 xn = table[x >> 10] << 1; 61 } 62 } 63 64 xn = (xn + 1 + (x / xn)) >> 1; 65 66 return ((xn * xn) > x) ? --xn : xn; 67 } 68 } else { 69 if (x >= 0x100) { 70 if (x >= 0x1000) { 71 if (x >= 0x4000) { 72 xn = (table[x >> 8]) + 1; 73 } else { 74 xn = (table[x >> 6] >> 1) + 1; 75 } 76 } else { 77 if (x >= 0x400) { 78 xn = (table[x >> 4] >> 2) + 1; 79 } else { 80 xn = (table[x >> 2] >> 3) + 1; 81 } 82 } 83 84 return ((xn * xn) > x) ? --xn : xn; 85 } else { 86 if (x >= 0) { 87 return table[x] >> 4; 88 } 89 } 90 } 91 92 return -1; 93 } 94 public static void main(String[] args){ 95 System.out.println(sqrt(65)); 96 97 } 98 }