leetcode解题方案--043-- Multiply Strings

题目

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.

Note:

The length of both num1 and num2 is < 110.
Both num1 and num2 contains only digits 0-9.
Both num1 and num2 does not contain any leading zero.
You must not use any built-in BigInteger library or convert the inputs to integer directly.

分析

两个可能狠狠狠大的数相乘。考察乘法的分解。如何分解乘法再把结果想加？
我用了递归，直到把乘法分解成一位数成一位数为止。

public static String multiply(String num1, String num2) {        char[] n1 = num1.toCharArray();        char[] n2 = num2.toCharArray();        if (n1.length == 1 && n2.length == 1) {            return String.valueOf((n1[0] - '0') * (n2[0] - '0'));        }        else if (n2.length == 1) {            long sum = 0;            for (int i = 0; i < n1.length; i++) {                sum = sum*10;                sum = sum + Integer.parseInt(multiply(num2,String.valueOf(num1.substring(i, i+1))));            }            return String.valueOf(sum);        }        else if (n1.length == 1) {            return multiply(num2, num1);        } else {            String sum = "";            for (int i = 0; i < n1.length; i++) {                sum = sum+"0";                String xx = multiply(num2,String.valueOf(num1.substring(i, i+1)));                char[] big = (sum.length()>xx.length()?sum:xx).toCharArray();                char[] small = (sum.length()<=xx.length()?sum:xx).toCharArray();                int add = 0;                for (int k = 0;k<small.length;k++) {                    int a = small[small.length-1-k]-'0';                    int b = big[big.length-1-k]-'0';                    int addd = a+b+add;                    big[big.length-1-k] = (char) (addd%10+'0');                    add = addd/10;                }                int p = big.length - small.length -1;                while (add!=0 && p>=0) {                    big[p] = (char) ((big[p]+add-'0')%10+'0');                    add=big[p]+add-'0';                    add = add/10;                    p--;                }                sum = new String(big);                if (add!=0) {                    sum ="1"+sum;                }            }            return String.valueOf(sum);        }    }

迄今以来惟一一次晒了ac details，想哭。。
简单分析了一下原因，主要还是拆箱装箱和string的拼接。
比较好的地方是中间运算用了数组，
比较渣的地方是 为了不想多写一个方法，全程string。

建议用stringbuffer来递归，只是需要重写一个递归函数。