题目描述

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.

分析

代码1

    public String multiply(String num1, String num2) {

        if (num1 == null || num2 == null) {
            return "";
        }

        int[] paper = new int[num1.length() + num2.length()];

        char[] _num1 = num1.toCharArray();
        char[] _num2 = num2.toCharArray();

        for (int i = 0; i < _num1.length; i++) {
            for (int j = 0; j < _num2.length; j++) {
                paper[paper.length - (i + j + 2)] += (_num1[i] - '0')
                        * (_num2[j] - '0');
            }
        }

        // add
        for (int i = 0; i < paper.length - 1; i++) {
            paper[i + 1] += paper[i] / 10;
            paper[i] %= 10;
        }

        String s = "";
        for (int i = paper.length - 1; i > 0; i--) {

            if ("" == s && paper[i] == 0) {
                continue;
            }
            s += paper[i];
        }

        s += paper[0];

        return s;
    }

代码2

使用Java的API：BigInteger，能对任意位数的整数进行数学基本运算。但是在LeetCode中不能AC。

    public String multiply2(String num1, String num2) {

        if (num1 == null || num2 == null) {
            return "";
        }

        BigInteger n1 = new BigInteger(num1);
        BigInteger n2 = new BigInteger(num2);

        return n1.multiply(n2).toString();
    }