题目描述

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

Example 1:

Input: num1 = “2”, num2 = “3”
Output: “6”
Example 2:

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.

算法1

(暴力枚举) O(n^2)

思路

最后把leading zero删除就可以了。

Java 代码

class Solution {

    public String multiply(String a, String b) {
        if (a == null || "".equals(a) || b == null || "".equals(b)) {
            return "0";
        }

        if ("0".equals(a) || "0".equals(b)) {
            return "0";
        }

        int n1 = a.length(), n2 = b.length();
        int[] products = new int[n1 + n2];
        for (int i = n1 - 1; i >= 0; i--) {
            for (int j = n2 - 1; j >= 0; j--) {
                int d1 = a.charAt(i) - '0';
                int d2 = b.charAt(j) - '0';
                products[i + j + 1] += d1 * d2;
            }
        }
        int carry = 0;
        for (int i = products.length - 1; i >= 0; i--) {
            int tmp = (products[i] + carry) % 10;
            carry = (products[i] + carry) / 10;
            products[i] = tmp;
        }
        StringBuilder sb = new StringBuilder();
        for (int num : products) {
            sb.append(num);
        }
        while (sb.length() != 0 && sb.charAt(0) == '0') {
            sb.deleteCharAt(0);
        }
        return sb.length() == 0 ? "0" : sb.toString();
    }

}