Leetcode-Multiply Strings

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.

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Solutiion:

a[i]*b[j] is the value of c[i+j]. So we have

 

Iter a[i]

  iter b[j]

      c[i+j] = a[i]*b[j]+c[i+j]

      carry = c[i+j]/10;

      c[i+j] = c[i+j]%10;

     c[i+j+1] += carry.  //NOTE: because we are dealing with string, we need to manipulate the char directly, after adding carry to c[i+j+1], c[i+j+1] may be larger than 10. Therefore, we need to keep dealing with carry, until in some higher bit, the carry becomes 0!

 

 Newest Solution:

Use int array, address all carry together afte calculation.

public class Solution {
    public String multiply(String num1, String num2) {
        if (num1.isEmpty() || num2.isEmpty()) return "";
        if (num1.equals("0") || num2.equals("0")) return "0";
        
        int len1 = num1.length(), len2 = num2.length();
        int[] res = new int[len1+len2];
        int[] n1 = new int[len1];
        int[] n2 = new int[len2];
        
        for (int i=0;i<len1;i++){
            n1[i] = getValue(num1.charAt(len1-1-i));
        }
        
        for (int i=0;i<len2;i++){
            n2[i] = getValue(num2.charAt(len2-1-i));
        }
        
        
        for (int j=0;j<len2;j++)
            for (int i=0;i<len1;i++){
                res[i+j] += n1[i]*n2[j];
            }
        
        StringBuilder builder = new StringBuilder();    
        for (int i=0;i<len1+len2-1;i++){
            res[i+1] += res[i] / 10;
            res[i] = res[i] % 10;
            builder.append(res[i]);
        }
        if (res[len1+len2-1]!=0)
            builder.append(res[len1+len2-1]);
        
        
        
        return builder.reverse().toString();
    }
    
    private static int getValue(char a){
        return (int) (a-'0');
    }
    
}

public class Solution {
    public String multiply(String num1, String num2) {
        if (num1.isEmpty() || num2.isEmpty()) return "";
        if (num1.equals("0") || num2.equals("0")) return "0";
        if (num1.equals("1")) return num2;
        if (num2.equals("1")) return num1;


        StringBuilder b1 = new StringBuilder(num1);
        StringBuilder b2 = new StringBuilder(num2);
        b1.reverse();
        b2.reverse();
        StringBuilder b3 = new StringBuilder();

        //perform multiplication bit by bit;
        for (int i=0;i<b1.length();i++){
            char c1 = b1.charAt(i);
            for (int j=0;j<b2.length();j++){
                char c2 = b2.charAt(j);
                int val = (c1-'0')*(c2-'0');
                if (b3.length()>(i+j)) val += b3.charAt(i+j)-'0';
                int carry = val/10;
                val = val % 10;
                //put val of current bit into string.
                if (b3.length()>(i+j))
                    b3.replace(i+j,i+j+1,Integer.toString(val));
                else b3.append((char)(val+'0'));

                //put carry into (i+j+1) bit.
                if (carry==0) continue;
                int curBit = i+j+1;
                while (carry!=0){ 
                    if (b3.length()>curBit){
                        val = (b3.charAt(curBit)-'0')+carry;
                        carry = val / 10;
                        val = val%10;
                        b3.replace(curBit,curBit+1,Integer.toString(val));
                        curBit++;
                    } else {
                        b3.append((char)(carry+'0'));
                        carry = 0;
                    }
                }
            }
        }

        b3.reverse();

        return b3.toString(); 
    }
}

 Solution2:

Directly use char arry. It is faster than StringBuilder.

public class Solution {
    public String multiply(String num1, String num2) {
        if (num1.isEmpty() || num2.isEmpty()) return "";
        if (num1.equals("0") || num2.equals("0")) return "0";
        if (num1.equals("1")) return num2;
        if (num2.equals("1")) return num1;


        StringBuilder temp = new StringBuilder(num1);
        temp.reverse();
        char[] b1 = temp.toString().toCharArray();
        temp = new StringBuilder(num2);
        temp.reverse();
        char[] b2 = temp.toString().toCharArray();
        char[] b3 = new char[b1.length+b2.length];
        Arrays.fill(b3,'0');


        //perform multiplication bit by bit;
        for (int i=0;i<b1.length;i++){
            char c1 = b1[i];
            for (int j=0;j<b2.length;j++){
                char c2 = b2[j];
                int val = (c1-'0')*(c2-'0') + (b3[i+j]-'0');
                int carry = val/10;
                val = val % 10;
                //put val of current bit into string.
                b3[i+j] = (char) (val+'0');

                //put carry into (i+j+1) bit.
                if (carry==0) continue;
                int curBit = i+j+1;
                while (carry!=0){ 
                    val = (b3[curBit]-'0')+carry;
                    carry = val / 10;
                    val = val%10;
                    b3[curBit] = (char)(val+'0'); 
                    curBit++;
                }
            }
        }
       
        temp = new StringBuilder();
        int index = b3.length-1;
        for (int i=b3.length-1;i>=0;i--) 
            if (b3[i]!='0'){
                index = i;
                break;
            }
        for (int i=index;i>=0;i--) temp.append(b3[i]);
        

        return temp.toString(); 
    }
}