Handling large integers in Java without using BigInteger

问题

I am currently working on an assignment that involves calculating Fibonacci numbers up to F(300) using a linear time complexity algorithm. I know via a quick search that a Java long will overflow at around F(93), so I am trying to figure out how to store such a large number.

However, our assignment states that we are not allowed to use language libraries such as BigInteger to store our numbers. We must write our own large integer data type or object.

What exactly am I to do in this case? I have never before had to write my own data type to handle something like a large integer. Is there some usual method of doing this?

回答1:

I believe what you need is to create a data type out of primitive arrays, like int[] or long[] so that you store result value by bits across elements of this array. Just like you can use two int's to store a long's bits you make take more ints to store a longer number.

One problem though is to interpret this value, say, you wish to print the value - that would require an algorithm on its own, if a string like 2^n1 + 2^n2 + ... + 2^nk is not acceptable.

An example (using byte array to store decimal digits):

import java.util.ArrayList;
import java.util.List;


public final class HumongousInt {
public static HumongousInt of( String string ) {
    return new HumongousInt( asByteArray( string ) );
}

private static byte[] asByteArray( String string ) {
    int length = string.length();
    byte[] data = new byte[ length ];
    for( int i = 0; i < length; i++ ) {
        data[ i ] = asByte( string.charAt( i ) );
    }
    return data;
}

private static byte asByte( char c ) {
    if( c > '9' && c < '0' ) {
        throw new IllegalArgumentException( "Wrong numbe format only numbers 0-9 could be present" );
    }
    return ( byte ) ( c - '0' );
}

private final byte[] digits;

private HumongousInt( byte[] digits ) {
    this.digits = digits;
}

public HumongousInt add( HumongousInt other ) {
    int maxLength = Math.max( this.digits.length, other.digits.length );
    ArrayList< Byte > data = new ArrayList<>();
    int overhead = 0;
    int i = 1;
    while( overhead > 0 || ( i <= maxLength ) ) {
        int currentDigit = overhead;
        int thisIndex = ( this.digits.length - i );
        if( thisIndex >= 0 ) {
            currentDigit += this.digits[ thisIndex ];
        }
        int otherIndex = ( other.digits.length - i );
        if( otherIndex >= 0 ) {
            currentDigit += other.digits[ otherIndex ];
        }
        overhead = currentDigit / 10;
        data.add( Byte.valueOf( ( byte ) ( currentDigit %= 10 ) ) );
        i++;
    }
    return new HumongousInt( asInvertedByteArray( data ) );
}

private byte[] asInvertedByteArray( List< Byte > list ) {
    byte[] data = new byte[ list.size() ];
    for( int i = data.length - 1, j = 0; i >= 0; i--, j++ ) {
        data[ j ] = list.get( i );
    }
    return data;
}

public HumongousInt add( int i ) {
    return add( HumongousInt.of( Integer.toString( i ) ) );
}

@Override
public String toString() {
    StringBuilder builder = new StringBuilder( digits.length );
    for( byte b : digits ) {
        builder.append( b );
    }
    return ( digits[ 0 ] == ( byte ) 0 ) ? builder.substring( 1 ) : builder.toString();
}
}

来源：https://stackoverflow.com/questions/21490086/handling-large-integers-in-java-without-using-biginteger

标签

java

long-integer

biginteger

fibonacci