解决算法;
*  1. 模拟小学乘法：最简单的乘法竖式手算的累加型；
*  2. 分治乘法：最简单的是Karatsuba乘法，一般化以后有Toom-Cook乘法；
*  3. 快速傅里叶变换FFT：（为了避免精度问题，可以改用快速数论变换FNTT），时间复杂度O(N lgN lglgN)。具体可参照Schönhage–Strassen algorithm；
*  4. 中国剩余定理：把每个数分解到一些互素的模上，然后每个同余方程对应乘起来就行；
*  5. Furer’s algorithm：在渐进意义上FNTT还快的算法。不过好像不太实用，本文就不作介绍了。大家可以参考维基百科Fürer’s algorithm

参考地址：https://blog.csdn.net/u010983881/article/details/77503519

1. 小学乘法代码如下：

/*** 模拟小学乘法*/public static String primaryMul(String str1, String str2){System.out.println("start primaryMul param=" + str1 + "," + str2 + ";time=" + System.currentTimeMillis());int[] int1 = string2IntArray(str1);int[] int2 = string2IntArray(str2);List<Integer> result = new ArrayList<>();for (int i = int1.length-1; i >=0 ; i--) {List<Integer> tempList = new ArrayList<>();int ten = 0;for (int j = int2.length-1; j >=0 ; j--) {int a = int1[i] * int2[j] + ten;ten = a / 10;tempList.add(a % 10);}if (ten > 0){tempList.add(ten);}//两行相加int carry = 0;for (int j = 0; j < tempList.size(); j++) {int r1 = 0;int index = j + (int1.length-1-i);if (index < 0 | result.size() > index){r1 = result.get(index);}int a = tempList.get(j) + r1 + carry;carry = a / 10;if (index < 0 | result.size() > index){result.set(index, a % 10);}else{result.add(a % 10);}}if (carry > 0){result.add(carry);}}//方向旋转int copy;for (int i = 0; i < result.size(); i++) {if (i < result.size() / 2){copy = result.get(i);int index = result.size() -1 -i;result.set(i, result.get(index));result.set(index, copy);}}StringBuilder sb = new StringBuilder();for (Integer integer : result) {sb.append(integer);}System.out.println("end primaryMul result=" + sb + ";time=" + System.currentTimeMillis());return sb.toString();}public static int[] string2IntArray(String str){String[] strs = str.split("");int[] ret = new int[strs.length];for (int i = 0; i < strs.length; i++) {ret[i] = Integer.parseInt(strs[i]);}return ret;}

2. 小学算法-累加算法

 

/*** 模拟小学乘法 - 累加算法*/private static String accumul(String str1, String str2){System.out.println("start accumul param=" + str1 + "," + str2 + ";time=" + System.currentTimeMillis());int[] int1 = string2IntArray(str1);int[] int2 = string2IntArray(str2);int[] result = new int[int1.length + int2.length];for (int i = 0; i < int1.length; i++) {for (int j = 0; j < int2.length; j++) {result[i + j + 1] += int1[i] * int2[j];}}for (int i = result.length - 1; i >= 0; i--) {if (result[i] >= 10){result[i-1] += result[i] / 10;result[i] = result[i] % 10;}}StringBuilder sb = new StringBuilder();for (Integer integer : result) {sb.append(integer);}System.out.println("end accumul result=" + sb + ";time=" + System.currentTimeMillis());return sb.toString();}

3. 分治算法-Karatsuba

/*** Karatsuba乘法*/
public static long karatsuba(long num1, long num2){//递归终止条件if(num1 < 10 || num2 < 10) return num1 * num2;// 计算拆分长度int size1 = String.valueOf(num1).length();int size2 = String.valueOf(num2).length();int halfN = Math.max(size1, size2) / 2;/* 拆分为a, b, c, d */long a = Long.valueOf(String.valueOf(num1).substring(0, size1 - halfN));long b = Long.valueOf(String.valueOf(num1).substring(size1 - halfN));long c = Long.valueOf(String.valueOf(num2).substring(0, size2 - halfN));long d = Long.valueOf(String.valueOf(num2).substring(size2 - halfN));// 计算z2, z0, z1, 此处的乘法使用递归long z2 = karatsuba(a, c);long z0 = karatsuba(b, d);long z1 = karatsuba((a + b), (c + d)) - z0 - z2;return (long)(z2 * Math.pow(10, (2*halfN)) + z1 * Math.pow(10, halfN) + z0);
}