上一篇写的“[大整数乘法]分治算法的时间复杂度研究”，这一篇是基于上一篇思想的代码实现，以下是该文章的连接：
http://www.cnblogs.com/McQueen1987/p/3348426.html
代码主要实现大整数乘法，过程中也涉及到[大整数加法] 和 [大整数减法] 的计算，代码如下：
类1
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package bigIntNum;
public class NumDividEqual {public char[] A;public char[] B;int n;/*** 将数组均分为两份，分别存入数组A和数组B中；* @param input*/public NumDividEqual(char[] input){n = input.length/2;A = new char[n];B = new char[n];for(int i = 0; i<n;i++){A[i] = input[i];}for(int i = 0; i<n;i++){B[i] = input[i + n];} }public static void main(String[] args) {// TODO Auto-generated method stub}
}
类2
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package bigIntNum;import java.util.Arrays;public class bigIntMult {/*** 将字符数组倒序排列* @param input* @return*/public char[] reverse(char[] input) {char[] output = new char[input.length];for (int i = 0; i < input.length; i++) {output[i] = input[input.length - 1 - i];}return output;}/*** 将大整数平均分成两部分* @param input* @return*/public NumDividEqual partition(char[] input) {return new NumDividEqual(input);}/*** 求两数组中较大数组的长度，如果其长度为奇数则+1变偶* @param num1* @param num2* @return*/public int calLength(char[] num1, char[] num2) {int len = num1.length > num2.length ? num1.length : num2.length;if (len == 1)return 1;len += len & 1;return len;}/*** 除去数字前面多余的0* @param input* @return*/public static char[] trimPrefix(char[] input) {char[] ret = null;for (int i = 0; i < input.length; i++) {if (ret == null && input[i] == '0')continue;else {if (ret == null) {ret = new char[input.length - i];//出去数字前面多余的0 }ret[i - (input.length - ret.length)] = input[i];}}if (ret == null)return new char[] { '0' };return ret;}/*** 数组如果长度不足n，则在数组前面补0，使长度为n。* @param input 输入数组要求数字的最高位存放在数组下标最小位置* @param n * @return*/public static char[] format(char[] input, int n) {//；if (input.length >= n) {return input;}char[] ret = new char[n];for (int i = 0; i < n - input.length; i++) {ret[i] = '0';}for (int i = 0; i < input.length; i++) {ret[n - input.length + i] = input[i];}return ret;}/*** 大整数尾部补0。相当于移位，扩大倍数* @param input* @param n* @return*/public char[] addTail(char[] input, int n) {// char[] ret = new char[input.length + n];for (int i = 0; i < input.length; i++) {ret[i] = input[i];}for (int i = input.length; i < ret.length; i++) {ret[i] = '0';}return ret;}/*** 大整数加法* @param num1* @param num2* @return*/public char[] add(char[] num1, char[] num2) {int len = num2.length > num1.length ? num2.length : num1.length;int carry = 0;//进位标识num1 = format(num1, len);num2 = format(num2, len);char[] ret = new char[len + 1];for (int i = len - 1; i >= 0; i--) {int tmp = num1[i] + num2[i] - 96;tmp += carry;if (tmp >= 10) {carry = 1;tmp = tmp - 10;} else {carry = 0;}ret[len - i - 1] = (char) (tmp + 48);}ret[len] = (char) (carry + 48);//最后一次，最高位的进位return trimPrefix(reverse(ret));}/*** 大整数减法：* @param num1 被减数，大整数乘法中只有一个减法(A+B)(C+D)-(AC+BD)=AC+BC>0,因此參數num1>num2且都为正* @param num2 减数* @return*/public static char[] sub(char[] num1, char[] num2) {int lenMax = num1.length > num2.length ? num1.length : num2.length;char[] newNum1 = Arrays.copyOf(format(num1, lenMax), lenMax);//字符串前面补0，使两串长度相同char[] newNum2 = Arrays.copyOf(format(num2, lenMax), lenMax);for(int i=0;i<lenMax;i++){//when num1-num2<0 returnif((newNum1[i]=='0' && newNum1[i]=='0') || newNum1[i] == newNum2[i]){//newNum1 is bigger; continue;}else if(newNum1[i] < newNum2[i]){//不滿足參數num1>num2；System.out.println("The Parameter in sub(A,B).A MUST Bigger Than B!");System.exit(0);}else break;}for(int i=lenMax-1;i>=0;i--){if(newNum1[i] < newNum2[i]){//result < 0newNum1[i] = (char) (newNum1[i] + '0' + 10 - newNum2[i]);newNum1[i-1] = (char) (newNum1[i-1] - 1);}else{newNum1[i] = (char) (newNum1[i] + '0' - newNum2[i]);}}return trimPrefix(newNum1);} /*** 大整数乘法* @param num1* @param num2* @return*/public char[] mult(char[] num1, char[] num2) {char[] A, B, C, D, AC, BD, AjB, CjD, ACjBD, AjBcCjD, SUM;int N = calLength(num1, num2);//求两数组中较大数组的长度，如果长度为奇数则+1变偶，方便二分成两部分num1 = format(num1, N);//数组高位存整数的高位数；数字前面补0，使长度为n；num2 = format(num2, N);if (num1.length > 1) {NumDividEqual nu1 = partition(num1);//将大整数平均分成两部分NumDividEqual nu2 = partition(num2);A = nu1.A;B = nu1.B;C = nu2.A;D = nu2.B; AC = mult(A, C);//分治求大整数乘法BD = mult(B, D);AjB = add(A,B);CjD = add(C,D);ACjBD = add(AC,BD);AjBcCjD = mult(AjB, CjD);char[] tmp1 = addTail(sub(AjBcCjD, ACjBD), N / 2);//尾部补0，相当于移位char[] tmp2 = add(addTail(AC, N), BD);SUM = add(tmp1, tmp2);char[] test = trimPrefix(SUM);//除去结果前面多余的0return test;} else {Integer ret = (num1[0] - 48) * (num2[0] - 48);return ret.toString().toCharArray();}}public static void main(String[] args) {String st1 = "168746315641347979798";String st2 = "164681654767446887797451316158";char[] a = st1.toCharArray();char[] b = st2.toCharArray();bigIntMult bg = new bigIntMult();char[] ret = bg.mult(a, b);System.out.println(ret);} }
代码优化：
1.可以写个hash表，存储一些常见的乘法，从而避免每次都重复计算。比如9999x9999999，里面有重复的9x9计算，可以考虑hash表存储这些计算的结果，用到时直接查询结果，从而避免重复计算。
声明：代码是本人脑力劳动结果，请尊重他人劳动。转载注明出处！