java实现大数相乘
计算大数：1234567891011121314151617181920 X 2019181716151413121110987654321？
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计算结果：2492816912877266687794240983772975935013386905490061131076320; length=61
验证结果：2.4928169128772666E60
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源代码如下：
/** * 计算大数相乘 * @author zhoujianghai * zhoujiangbohai@163.com */public class BigNumberCount { public static void main(String agrs[]){ String number1 = "1234567891011121314151617181920"; String number2 = "2019181716151413121110987654321"; String result = multipBigNumber(number1,number2); System.out.println("计算结果："+result+"; length="+result.length()); System.out.println("验证结果："+Double.parseDouble(number1) * Double.parseDouble(number2)); } public static String multipBigNumber(String s1,String s2){ int longArray[] = null; int shortArray[] = null; int s1Length = s1.length(); int s2Length = s2.length(); int longLength = s1Length > s2Length ? s1Length : s2Length; int shortLength = s1Length == longLength ? s2Length : s1Length; longArray = new int[longLength]; shortArray = new int[shortLength]; String longString = s1Length >= s2Length ? s1 : s2; String shortString = s1.equals(longString) ? s2 : s1; System.out.println("longString="+longString+"; shortString="+shortString); //低位在前，高位在后 for(int i = longLength - 1; i >= 0; i --){ longArray[longLength - 1 - i] = longString.charAt(i) - 48; } for(int i = shortLength - 1; i >= 0; i --){ shortArray[shortLength - 1 - i] = shortString.charAt(i) - 48; } StringBuffer results[] = new StringBuffer[longLength]; for(int i = 0; i < results.length; i++) { results[i] = new StringBuffer(); } StringBuffer resultBuffer = new StringBuffer(); /** * 把被乘数的每一位与乘数逐位相乘 * 如：5607 * 2256,被乘数5607，乘数2256 * 7 * 6 + 0 = 42 ：0表示低位向高位的进位 7 * 5 + 4 = 39 7 * 2 + 3 = 17 7 * 2 + 1 = 15 29751 0 * 6 + 0 = 0 0 * 5 + 0 = 0 0 * 2 + 0 = 0 0 * 2 + 0 = 0 00000 6 * 6 + 0 = 36 6 * 5 + 3 = 33 6 * 2 + 3 = 15 6 * 2 + 1 = 13 63531 5 * 6 + 0 = 30 5 * 5 + 3 = 28 5 * 2 + 2 = 12 5 * 2 + 1 = 11 08211 * results：29751， 00000，63531，08211 * */ for(int i = 0; i < longLength; i ++){ int temp = 0; int tempCarry = 0;//低位向高位的进位 int currentValue = 0; //当前位乘积的值 for(int j = 0; j < shortLength; j ++){ temp = longArray[i] * shortArray[j] + tempCarry; //System.out.println("longArray[i]="+longArray[i]+" * "+"shortArray[j]="+shortArray[j]+" + "+tempCarry+" = "+temp); tempCarry = temp / 10; currentValue = temp % 10; results[i].append(currentValue); if(j == shortLength - 1){ results[i].append(tempCarry);//最高位有进位则进位，无进位则补0 } } System.out.println( results[i].toString()); } /**低位补0，results[i]低位补i个0(低位在前，高位在后) * 29751000000006353100008211 * */ for(int i = 0,length = results.length;i < length; i++){ String temp = ""; for(int j = 0; j < i; j ++){ temp += "0"; } results[i].insert(0, temp); System.out.println( results[i].toString()); } /**把每一行相加(低位在前，高位在后,从左往右 加) * 29751000000006353100008211 ------------- 29394621 * * */ int tempCarry = 0; int currentValue = 0; StringBuffer lastBuffer = results[results.length - 1]; //需要循环的次数 int times = lastBuffer.length(); for(int i = 0; i < times; i ++ ){ int temp = 0; for(int j = 0,length2 = results.length; j < length2; j ++){ if(results[j].length() > i){ temp += (results[j].charAt(i) - 48);//把第j行的第i位的字符转换成int } } temp += tempCarry; tempCarry = temp / 10; currentValue = temp % 10; resultBuffer.append(currentValue); } StringBuffer finalResult = new StringBuffer(); boolean hasNumNotZero = false; //因为低位在前，高位在后，所以把29394621 转换成最终结果：12649392 for(int i = resultBuffer.length() - 1; i >= 0; i --) { int temp = resultBuffer.charAt(i) - 48; if(temp != 0) hasNumNotZero = true; if (hasNumNotZero) finalResult.append(temp); } if (finalResult.length() == 0) finalResult.append(0); return finalResult.toString(); } }?
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1 楼 faulware 2012-06-25 本以为这么长的代码 肯定考虑了结果位数的问题，结果.... StringBuffer results[] = new StringBuffer[longLength]; 那么13*9结果是2位数吗？？？ 2 楼 zhoujianghai 2012-06-25 faulware 写道本以为这么长的代码 肯定考虑了结果位数的问题，结果.... StringBuffer results[] = new StringBuffer[longLength]; 那么13*9结果是2位数吗？？？