LeetCode | 43. Multiply Strings

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博客给出一道Leetcode题目，要求计算两个以字符串表示的非负整数的乘积，结果也用字符串表示。题目给出示例和注意事项，如数字长度小于110、仅含数字、不能用内置大整数库等，还给出了代码。

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题目：

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 integerdirectly.

代码：

class Solution {
public:
    string multiply(string num1, string num2) {
        if (num1 < num2)
			swap(num1, num2);
		if (num2 == "0")
			return "0";

		string res = "";
		vector<int> val(num1.length() + num2.length() + 1, 0);

		reverse(num1.begin(), num1.end());
		reverse(num2.begin(), num2.end());
		for (int i = 0; i < num2.length(); i++)
		{
			if (num2[i] == '0')
				continue;
			vector<int> tmp_val(num1.length()+1, 0);
			for (int j = 0; j < num1.length(); j++)
			{
				int cur2 = num2[i] - '0';
				int cur1 = num1[j] - '0';
				tmp_val[j] += (cur2 * cur1);
				if (tmp_val[j] > 9)
				{
					tmp_val[j + 1] = tmp_val[j] / 10;
					tmp_val[j] %= 10;
				}
			}
			for (int j = 0; j < tmp_val.size(); j++)
			{
				val[i + j] += tmp_val[j];
				if (val[i + j] > 9)
				{
					val[i + j + 1] += val[i + j] / 10;
					val[i + j] %= 10;
				}
			}
		}
		int i = val.size() - 1;
		for (; i >= 0 && val[i] == 0; i--);
		for (; i >= 0; i--)
		{
			char cur_val = val[i] + '0';
			res += cur_val;
		}
		return res;
    }
};

题外话：

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