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.

public class Solution {

    //模拟手算乘法

    public String multiply(String num1, String num2) {

        int n = num2.length();

        String[] addS = new String[n];

        for(int i = 0; i < n; i++){

            addS[i] = multiplyChar(num1, num2.charAt(i), n-1-i);

        }

        String res = sum(addS);

        int p = 0;

        while(p < res.length() && res.charAt(p) == '0')

        	p++;

        if(p == res.length())

        	return "0";

        else

        	return res.substring(p);

    }

    public String multiplyChar(String num, char c, int digits){

        int n = num.length();

        char[] res = new char[n + 1];

        int add = 0;

        for(int i = n; i >= 0; i--){

            int b = 0;

            if(i-1 >= 0)

                b = num.charAt(i-1)-'0';

            int cur = b*(c-'0')+add;

            add = cur / 10;

            cur %= 10;

            res[i] = (char)(cur+'0');

        }

        int p = 0;

        while(p <= n && res[p] == '0')

            p++;

        if(p == n + 1)

        	return "0";

        else{

            StringBuffer sb = new StringBuffer();

            for(int i = 0; i < digits; i++){

                sb.append('0');

            }

        	return new String(res, p, n-p+1) + sb.toString();

        }

    }

    public String sum(String[] s){

        StringBuffer res= new StringBuffer();

        int p = 0;

        int cur = 0;

        int add = 0;

        int maxLength = 0;

        for(int i = 0; i < s.length; i++){

            maxLength = Math.max(maxLength, s[i].length());

        }

        do{

        	cur = 0;

            for(int i = 0; i < s.length; i++){

                if(p < s[i].length())

                    cur += s[i].charAt(s[i].length()-1-p)-'0';

            }

            cur += add;

            res.append(cur%10);

            add = cur / 10;

            p++;

        }

        while(cur != 0 || p < maxLength);

        return new String(res.reverse());

    }

}