• Java实现大整数乘法


    1 问题描述
    计算两个大整数相乘的结果。

    2 解决方案
    2.1 蛮力法

    package com.liuzhen.chapter5;
    
    import java.math.BigInteger;
    
    public class BigNumber {
        /*
         * 参数A:进行乘法运算的大整数A,用字符串形式表示
         * 参数B:进行乘法运算的另一个大整数B,用字符串形式表示
         * 函数功能:以字符串形式返回A*B的结果
         */
        public String getMultiBigNumber(String A,String B){
            if(A.length() > B.length()){       //当B字符串长度小于A时,在B字符串前补0,使得两个字符串长度一致
                char[] temp = new char[A.length()-B.length()];
                for(int i = 0;i < A.length() - B.length();i++)
                    temp[i] = '0';
                B = String.valueOf(temp) + B;
            }
            if(A.length() < B.length()){      //当A字符串长度小于B时,在A字符串前补0,使得两字符串长度一致
                char[] temp = new char[B.length()-A.length()];
                for(int i = 0;i < B.length() - A.length();i++)
                    temp[i] = '0';
                A = String.valueOf(temp) + A;
            }
            
            int len = A.length() + B.length();
            
            char[] arrayA = A.toCharArray();
            char[] arrayB = B.toCharArray();
            for(int i = 0;i < arrayA.length;i++)     //检查字符串A中是否有非数字的字符
                if(arrayA[i] < '0' || arrayA[i] > '9')
                    return null;
            for(int i = 0;i < arrayB.length;i++)    //检查字符串B中是否有非数字的字符
                if(arrayB[i] < '0' || arrayB[i] > '9') 
                    return null;
            
            char[] result = new char[len];    //用于存放最终乘法运算结果,长度len表示A*B的最长长度
            for(int i = 0;i < len;i++)         //初始化字符数组result,各个元素均为'0'
                result[i] = '0';
            
            int countI = 0;       //用于计算当前B中已经和A中每个字符进行完乘法运算的字符个数    
            for(int i = arrayB.length-1;i >= 0;i--){
                int tempB = arrayB[i] - '0';
                int countJ = 0;   //用于计算当前A中正在进行乘法运算的字符个数
                for(int j = arrayA.length - 1;j >= 0;j--,countJ++){
                    int tempA = arrayA[j] - '0';
                    int tempRe = (tempB * tempA) % 10;  //用于计算当前位置的数
                    int tempResult = result[(len-1-countJ)-countI] - '0';  //当前位置已包含的结果
                    tempResult += tempRe;
                    //count--表示当前A字符串中进行乘法运算的字符位置,countI表示当前B字符串中进行乘法运算的字符位置
                    //(count--)-countI则表示当前进行乘法运算两个数字结果的最低位的位置
                    result[(len-1-countJ)-countI] = (char) (tempResult%10 + 48); //当前位置数最终结果
                    
                    int tempDi = tempB * tempA / 10 + tempResult / 10;       //用于计算进位
                    for(int k = 1;tempDi > 0;k++){   //处理进位操作
                         //当前下第k个位置包含的结果
                        int tempResultK = result[(len-1-countJ)-countI-k] - '0'; 
                        tempResultK += tempDi;
                        result[(len-1-countJ)-countI-k] = (char) (tempResultK%10 + 48);
                        tempDi = tempResultK / 10;
                    }
                }
                countI++;
            }
            
            return getNoneZeroString(result);
        }
        
        //去掉字符串前面的0
        public String getNoneZeroString(char[] result){
            int count = 0;
            for(int i = 0;i < result.length;i++){
                if(result[i] == '0')
                    count++;
                else
                    break;
            }
            char[] A = new char[result.length-count];
            for(int i = 0;i < result.length-count;i++)
                A[i] = result[count+i];
            return String.valueOf(A);
        }
        
        public static void main(String[] args){
            long t1 = System.currentTimeMillis();
            BigNumber test = new BigNumber();
            String A = "123456789123232342432423441345342523452534235443253254";
            String B = "987654322234242424332423414324532542354325235345435435";
            System.out.println("大整数A*B的结果:"+test.getMultiBigNumber(A, B));
            BigInteger bigInteger1 = new BigInteger("123456789123232342432423441345342523452534235443253254");
            BigInteger bigInteger2 = new BigInteger("987654322234242424332423414324532542354325235345435435");
            bigInteger2 = bigInteger2.multiply(bigInteger1);
            System.out.println("验证后A*B的结果:"+bigInteger2);
            long t2 = System.currentTimeMillis();
            System.out.println("耗时:"+(t2-t1)+" 毫秒");
        }
    }
    

    运行结果:

    大整数A*B的结果:121932631386721831198089710747298668585104317165230580938992491445929653074852215402191571860797295610655490
    验证后A*B的结果:121932631386721831198089710747298668585104317165230580938992491445929653074852215402191571860797295610655490
    耗时:4 毫秒
    
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  • 原文地址:https://www.cnblogs.com/a1439775520/p/12948049.html
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