Big integer multiplication is actually very simple (Java)

Big integer multiplication is actually very simple (Java)

Purpose and introduction

• Use Java language to achieve large integer multiplication
• As long as the big integers are multiplied, the addition, subtraction and division of big integers are very simple
• Add large integers, only need a loop for carry operation
• Subtract large integers, only need to borrow in the loop, pay attention to the sign.
• Dividing large integers is simply subtracting large integers multiple times.

Design ideas

• Simulate mathematical multiplication, multiply by bit, and add at the end
• When adding, if the low bit exceeds 10, move forward

Specific steps

• The operator is judged first, and the final result is positive or negative.
• Manual operations can be forwarded, but the computer performs string inversion, which is convenient for calculation and storage
• For the result of the calculation, you need to find a place to store it. The size of the new array is the number of digits of the product of the two numbers, which will not be greater than the sum of the two digits.
• Perform calculations and numerical processing (carry)
• Remember to remove the extra 0, and bring the sign

Code

public class BigIntegersMulti {
    
    
    //大整数乘法
    public String multi(String args, String args2) {
    
    
        StringBuffer sb = new StringBuffer(args);
        StringBuffer sb2 = new StringBuffer(args2);
        //运算符判断
        int signal = 0;
        if ("-".equals(args.substring(0, 1))) {
    
    
            sb.delete(0, 1);
            signal++;
        }
        if ("-".equals(args2.substring(0, 1))) {
    
    
            sb2.delete(0, 1);
            signal++;
        }
        //字符串逆置，方便计算和存储
        char[] ch = sb.reverse().toString().toCharArray();
        char[] ch2 = sb2.reverse().toString().toCharArray();
        //两个数相乘的积位数不会大于两数位数之和
        int length = ch.length + ch2.length;
        int[] sum = new int[length];
        //运算
        for (int i = 0; i < ch.length; i++) {
    
    
            for (int j = 0; j < ch2.length; j++) {
    
    
                sum[i + j] += (ch[i] - '0') * (ch2[j] - '0');
            }
        }
        //数值整理和进位
        for (int i = 0; i < length - 1; i++) {
    
    
            sum[i + 1] += (sum[i] / 10);
            sum[i] %= 10;
        }
        StringBuffer sb3 = new StringBuffer();
        //消除0，比如000123，应该输出123
        int u = length - 1;
        while (sum[u] == 0) {
    
    
            u--;
        }
        //逆序
        for (; u >= 0; u--) {
    
    
            sb3.append(sum[u]);
        }
        //带上运算符号
        if ((signal & 1) == 1) {
    
    
            sb3.insert(0, "-");
        }
        return sb3.toString();
    }
}

Test case design

• Test negative number multiplied by negative number
• Test positive numbers multiplied by negative numbers
• By 99 99,9999 give multiples of 9999 9 mathematical relationship, for multiplying the number of authentication (two identical 9 composed of the n, the result is the n-1 9 8 plus one plus the n-1 0 plus a 1)

public void testmulti() {
    
    

        //测试负数乘以负数
        String s1 = "-12345679", s2 = "-12345679";
        String s3 = bigIntegersMulti.multi(s1, s2);
        assert s3.equals("152415789971041");
        //测试正数乘以负数
        s1 = "12345679";
        s2 = "-12345679";
        s3 = bigIntegersMulti.multi(s1, s2);
        assert s3.equals("-152415789971041");
        //测试9的倍数关系
        StringBuffer sb = new StringBuffer();
        StringBuffer sb1 = new StringBuffer();
        sb.append(9);
        for (int i = 1; i < 20; i++) {
    
    
            sb.append(9);
            s3 = bigIntegersMulti.multi(sb.toString(), sb.toString());
            for (int j = 0; j < i; j++) {
    
    
                sb1.append(9);
            }
            sb1.append(8);
            for (int j = 0; j < i; j++) {
    
    
                sb1.append(0);
            }
            sb1.append(1);
            assert s3.equals(sb1.toString());
            sb1 = new StringBuffer();
        }
    }

Pit details

• Forgot to add sign
• No elimination 0