大整数乘法

１. 采用递归分治法

ab*cd=ac(bc+ad)bd

public String BigMutiple(String t1, String t2) {// 临界条件if (t1.length() <= size && t2.length() <= size) {return String.valueOf(Integer.parseInt(t1) * Integer.parseInt(t2));}// 预处理　make sure：t1.length = t2.lengthif (t1.length() != t2.length()) {boolean taglenth = t1.length() > t2.length() ? true : false;int distance = Math.abs(t1.length() - t2.length());if (taglenth) {t2 = belonger(t2, distance);} else {t1 = belonger(t1, distance);}}//ensureLength(t1, t2);String[] AB = numSplite(t1);String[] CD = numSplite(t2);String a = AB[0];String b = AB[1];String c = CD[0];String d = CD[1];int saveprefix = a.length();int savesuffix = b.length();assert (a.length() == c.length());assert (c.length() == d.length());String ac = BigMutiple(a, c);String ad = BigMutiple(a, d);String bc = BigMutiple(b, c);String bd = BigMutiple(b, d);return intergration(ac, ad, bc, bd, saveprefix, savesuffix);}private String belonger(String t1, int length) {StringBuilder sb = new StringBuilder("");for (int i = 0; i < length; i++) {sb.append("0");}return sb.toString() + t1;//如果使用字符数组，return charArray.toString()，返回的是对象名+hash地址一类的东西。}private String intergration(String ac, String ad, String bc, String bd, int saveprefix, int savesuffix) {String result = "";String[] endStrings = resultSplite(bd, savesuffix);assert (endStrings.length == 2);String endoverflow = endStrings[0];int savemiddle = (saveprefix >= savesuffix) ? saveprefix : savesuffix;//String[] middleStrings = resultSplite(Bigadd(ad, bc, endoverflow), savemiddle);String[] middleStrings = resultSplite(MutileAdd(ad, bc, endoverflow), savemiddle);String middleroverflow = middleStrings[0];String[] startStrings = resultSplite(Bigadd(ac, middleroverflow, "0"), saveprefix);result = startStrings[0] + startStrings[1] + middleStrings[1] + endStrings[1];return result;}private String Bigadd(String ad, String bc, String endoverflow) {// 将ad,bc,endoverflow相加String resultString = "";String result = "";if (ad.length() != bc.length()) {boolean taglenth = ad.length() > bc.length() ? true : false;int distance = Math.abs(ad.length() - bc.length());if (taglenth) {bc = belonger(bc, distance);} else {ad = belonger(ad, distance);}}assert (ad.length() == bc.length());int overtag = 0;for (int i = ad.length() - 1; i >= 0; i--) {int tempresult = (ad.charAt(i) - '0') + (bc.charAt(i) - '0') + overtag;if (tempresult < 10) {//bugresultString = (tempresult + "")+resultString;overtag = 0;//bug} else {resultString =  ((tempresult - 10) + "")+resultString;overtag = 1;}}//bug:最高为进位if (overtag!=0) {resultString=overtag+""+resultString;}if (resultString.length() != endoverflow.length()) {boolean taglenth = resultString.length() > endoverflow.length() ? true : false;int distance = Math.abs(resultString.length() - endoverflow.length());if (taglenth) {endoverflow = belonger(endoverflow, distance);} else {resultString = belonger(resultString, distance);}}//标志位的置位和恢复overtag = 0;for (int i = resultString.length() - 1; i >= 0; i--) {int tempresult = (resultString.charAt(i) - '0') + (endoverflow.charAt(i) - '0') + overtag;if (tempresult < 10) {//bugresult = (tempresult + "")+result ;overtag = 0;//bug} else {result =((tempresult - 10) + "")+ result;overtag = 1;}}//bug:最高位进位if (overtag!=0) {result=overtag+""+result;}return result;}private String MutileAdd(String anum,String bnum,String overbit){//make sure length is the same//maxlenth这个函数，返回值的含义预想的跟实际实现时是不同的。关键在于当时没有明确的确定。//maxlenth这个函数，返回最大的字符串的长度int maxlenth = maxlenth(anum,bnum,overbit);anum = belonger(anum, maxlenth-anum.length());bnum = belonger(bnum, maxlenth-bnum.length());overbit = belonger(overbit, maxlenth-overbit.length());assert(anum.length()==bnum.length());assert(bnum.length()==overbit.length());String resultString = "";int overflag = 0;for (int i = maxlenth-1; i>=0; i--) {//这里是３个数相加，进位有三种可能：0,1,2 所以用boolean不能表示int temp = (anum.charAt(i)-'0')+(bnum.charAt(i)-'0')+(overbit.charAt(i)-'0')+overflag;overflag = temp/10;temp = temp-overflag*10;resultString = (temp+"")+resultString;}if (overflag!=0) {resultString = overflag+resultString;}return resultString;}private int maxlenth(String anum, String bnum, String overbit) {// TODO Auto-generated method stubif (anum.length()>=bnum.length()) {if (anum.length()>=overbit.length()) {return anum.length();} else {return overbit.length();}} else {if (bnum.length()>=overbit.length()) {return bnum.length();} else {return overbit.length();}}}private String[] resultSplite(String bd, int savesuffix) {//assert (bd.length() >= savesuffix);if (bd.length()==savesuffix) {return new String[]{"0",bd};}else if (bd.length()>savesuffix) {//bug :在截取时，截取位置容易出错return new String[] { bd.substring(0, (bd.length()-savesuffix)), bd.substring(bd.length()-savesuffix) };}else{bd =belonger(bd, savesuffix-bd.length());return new String[]{"0",bd};}}private String[] numSplite(String t1) {int n = t1.length() / 2;String[] parts = { t1.substring(0, n), t1.substring(n) };return parts;}

2.不考虑进位的列式乘法

原理：

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public String twoMulti(String aString,String bString){//if (aString.length()<=size&&bString.length()<=size) {//return String.valueOf(Integer.parseInt(aString)*Integer.parseInt(bString)); //}//初始化为0int []result = new int[resultlenth];for (int i = aString.length()-1; i >=0; i--) {for (int j = bString.length()-1; j >=0; j--) {int temp = (aString.charAt(i)-'0')*(bString.charAt(j)-'0');int offset = aString.length()-1-i+bString.length()-1-j;result[offset]=result[offset]+temp;}}//对结果数组进行处理for (int i = 0; i < result.length-1; i++) {int save = result[i]%10;int overflow = result[i]/10;result[i]=save;result[i+1]+=overflow;}StringBuilder sBuilder = new StringBuilder("");boolean tag = true;for (int i = resultlenth-1; i >=0; i--) {if (result[i]==0&&tag) {continue;//i--;}else {tag=false;sBuilder.append(result[i]);}}return sBuilder.toString();}