043 Multiply Strings

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

Example：

Input: num1 = "2", num2 = "3"
Output: "6"

Input: num1 = "123", num2 = "456"
Output: "56088"

Note：

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

解释下题目：

两个大数相乘，记得不要使用内置的大数类

1. 老老实实做呗

实际耗时：19ms

public String multiply(String num1, String num2) {
        StringBuilder stringBuilder = new StringBuilder();
        int m = num1.length();
        int n = num2.length();
        if (m == 0 || n == 0 || num1.equals("0") || num2.equals("0")) {
            return "0";
        }
        int[] res = new int[m + n];

        for (int i = m - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                //两个对应位置的相乘
                int mul = (num1.charAt(i) - '0') * (num2.charAt(j) - '0');
                //和还要加上之前进位的(如果有的话)
                int sum = mul + res[i + j + 1];
                //填入
                res[i + j] += sum / 10;
                res[i + j + 1] = (sum) % 10;
            }
        }

        for (int i : res) {
            if (stringBuilder.length() == 0 && i == 0) {
                //去除开头的0
                continue;
            }
            stringBuilder.append(i);
        }

        return stringBuilder.toString();
    }

  思路：首先需要明确的一点是，一个m位的数和一个n位的数相乘得到的最大解是m+n位的，其次是第i位和第j位的两个数字相乘影响的是第i+j+i位，如果有进位就会影响它前面一位，也就是i+j位。弄懂这些之后去处理上面的这些就很简单了。

时间复杂度O(mn) m和n分别是两个String的长度

空间复杂度O(m+n) m和n分别是两个String的长度

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