几种常见求解平方根的方法

        //精度
	private final static double accuracy= 1e-6;

	/**
	 * 暴力求解
	 */
	public static double bruteSqrt(double x)
	{
		assert(x>=0);
		double ans=0.0;
		while (Math.abs(x - ans * ans) > accuracy)ans += accuracy;
		return ans;
	}

	/**
	 * 牛顿法求解
	 */
	public static double newtonSqrt(double x)
	{
		assert(x>=0);
		double avg = x;
		double last_avg = Double.MAX_VALUE;

		while (Math.abs(avg - last_avg) > accuracy)
		{
			last_avg = avg;
			avg = (avg + x / avg) / 2;
		}
		return avg;
	}

	/**
	 * 二分法求解
	 */
	public static double binarySqrt(double x)
	{
		assert(x>=0);

		double low = 0;
		double high = x;
		double mid = Double.MAX_VALUE;
		double last_mid = Double.MIN_VALUE;

		while (Math.abs(mid - last_mid) > accuracy)
		{
			last_mid = mid;
			mid = (low + high)/2;
			if (mid*mid>x)high = mid;
			if (mid*mid<x)low = mid;
		}
		return mid;

	}

	private final static int[] LUT =
	{ 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 };

	/**
	 * 查表法求解
	 */
	public static int intLutSqrt(int x)
	{
		int xn;

		if (x >= 0x10000)
		{
			if (x >= 0x1000000)
			{
				if (x >= 0x10000000)
				{
					if (x >= 0x40000000)
					{
						xn = LUT[x >> 24] << 8;
					}
					else
					{
						xn = LUT[x >> 22] << 7;
					}
				}
				else
				{
					if (x >= 0x4000000)
					{
						xn = LUT[x >> 20] << 6;
					}
					else
					{
						xn = LUT[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 = LUT[x >> 16] << 4;
					}
					else
					{
						xn = LUT[x >> 14] << 3;
					}
				}
				else
				{
					if (x >= 0x40000)
					{
						xn = LUT[x >> 12] << 2;
					}
					else
					{
						xn = LUT[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 = (LUT[x >> 8]) + 1;
					}
					else
					{
						xn = (LUT[x >> 6] >> 1) + 1;
					}
				}
				else
				{
					if (x >= 0x400)
					{
						xn = (LUT[x >> 4] >> 2) + 1;
					}
					else
					{
						xn = (LUT[x >> 2] >> 3) + 1;
					}
				}

				return ((xn * xn) > x) ? --xn : xn;
			}
			else
			{
				if (x >= 0)
				{
					return LUT[x] >> 4;
				}
			}
		}

		return -1;
	}

	/**
	 * Quake III中快速求解平方根倒数的方法
	 */
	public static float fastInvSqrt(float x)  
	{  
	     float xhalf = 0.5f*x;  
	     int f2i = Float.floatToRawIntBits(x);
	     f2i = 0x5f375a86-(f2i>>1);
	     x = Float.intBitsToFloat(f2i);
	     x = x*(1.5f-xhalf*x*x);
	     x = x*(1.5f-xhalf*x*x);
	     return x;  
	}
	
	/**
	 * Quake III中快速求解平方根方法
	 */
	public static float fastSqrt(float x) {  
	    float y=x;
	    float xhalf = 0.5f*x;
	    int f2i = Float.floatToRawIntBits(x);  
	    f2i = 0x5f3759df-(f2i>>1);  
	    x = Float.intBitsToFloat(f2i);  
	    x  = x * (1.5f-(xhalf*x*x));  
	    x  = x * (1.5f-(xhalf*x*x));  
	    return y*x;  
	}

	public static void main(String[] args)
	{
		System.out.println(bruteSqrt(3));
		System.out.println(newtonSqrt(3));
		System.out.println(binarySqrt(3));
		System.out.println(intLutSqrt(64));
		System.out.println(1/fastInvSqrt(3));
		System.out.println(fastSqrt(3));
	}