Leetcode:multiply Strings. Java

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

Note:the numbers can be arbitrarily large and is non-negative.

public class Solution {//Analog hand multiplication 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);        The 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].l            Ength () -1-p)-' 0 ';            } cur + = add;            Res.append (CUR%10);            add = CUR/10;        p++;        } while (cur! = 0 | | p < maxLength);    return new String (Res.reverse ()); }}

Leetcode:multiply Strings. Java