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());    }}

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