BigInteger开方运算

import java.math.BigInteger;
import java.util.Arrays;

/**
 * Created by MGL on 2017/4/21.
 */
public class BigInteger_Sqrt{
    public static void main(String[] args) {
        String str = "846516548651346132456465132189465134894651645613";
        BigInteger n1 = new BigInteger(str);
        BigInteger multiply = n1.multiply(n1);
        int length = multiply.toString().length();
        System.out.println("数的长度 = " + length);
        long start = System.nanoTime();
        BigInteger sqrt = getSqrt(multiply);
        long end = System.nanoTime();
        System.out.println(end - start);
        System.out.println("对" +length +"位数进行开放运算所需要的时间 = " + (end - start)/1000000000F + "s");
        System.out.println("运算结果 = " + (sqrt.compareTo(n1) == 0));
    }

    private static BigInteger getSqrt(BigInteger num) {
        String s = num.toString();
        int mlen = s.length();    //被开方数的长度
        int len;    //开方后的长度
        BigInteger beSqrtNum = new BigInteger(s);//被开方数
        BigInteger sqrtOfNum;    //存储开方后的数
        BigInteger sqrtOfNumMul;    //开方数的平方
        String sString;//存储sArray转化后的字符串
        if (mlen % 2 == 0) len = mlen / 2;
        else len = mlen / 2 + 1;
        char[] sArray = new char[len];
        Arrays.fill(sArray, '0');//开方数初始化为0
        for (int pos = 0; pos < len; pos++) {
            //从最高开始遍历数组，
            //每一位都转化为开方数平方后刚好不大于被开方数的程度
            for (char ch = '1'; ch <= '9'; ch++) {
                sArray[pos] = ch;
                sString = String.valueOf(sArray);
                sqrtOfNum = new BigInteger(sString);
                sqrtOfNumMul = sqrtOfNum.multiply(sqrtOfNum);
                if (sqrtOfNumMul.compareTo(beSqrtNum) == 1) {
                    sArray[pos] -= 1;
                    break;
                }
            }
        }
        return new BigInteger(String.valueOf(sArray));
    }
}

该算法的最坏情况是将循环中所有的数字都遍历完，为 len * 9，因此时间复杂度为 O(n)。