leetcode 43. Multiply Strings

最新推荐文章于 2024-04-28 07:28:07 发布

原创最新推荐文章于 2024-04-28 07:28:07 发布 · 156 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

280 篇文章

订阅专栏

本文介绍了一种模拟手工计算方式实现两个大整数相乘的算法。通过将输入的字符串形式的大整数逐位相乘并累加，最终得到乘积结果。此方法避免了直接转换成整数可能带来的溢出问题。

43. Multiply Strings

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.

模拟了小学学的那种两个数相乘的方式，按位相乘再相加。

class Solution {
public:
	string multiply(string num1, string num2)
	{
		string m1;
		if (num1.size() < num2.size()) //num2为长度短的那个
		{
			m1 = num1;
			num1 = num2;
			num2 = m1;
		}
        m1 = "";
		vector<vector<int>> result;
		vector<int> kk;
		int tt = 0;
		for (int i = num2.size() - 1; i >= 0; i--)
		{
			for (int h = i; h < num2.size() - 1; h++)
				kk.push_back(0);

			for (int j = num1.size() - 1; j >= 0; j--)
			{
				kk.push_back( ((int(num2[i]) - 48) * (int(num1[j]) - 48) + tt) % 10);
				tt = ((int(num2[i]) - 48) * (int(num1[j]) - 48) + tt) / 10;
			}
			if (tt != 0)
			{
				kk.push_back(tt);
				tt = 0;
			}
			result.push_back(kk);
			kk.clear();
		}

		int max = 0;
		for (int i = 0; i < result.size(); i++)
		{
			if (result[i].size() > max)
				max = result[i].size();
		}

		for (int i = 0; i < result.size(); i++)
		{
			tt = result[i].size();
			for (int p = tt; p < max; p++)
				result[i].push_back(0);
		}

		tt = 0;
		int flag = 0;
		for (int p = 0; p < max; p++)
		{
			tt = flag;
			for (int i = 0; i < result.size(); i++)
			{
				tt = result[i][p] + tt;
			}
			kk.push_back(tt % 10);
			flag = tt / 10;
		}
		if (flag != 0)
		{ 
			kk.push_back(flag);
			flag = 0;
		}
		for (int i = kk.size() - 1; i >= 0; i--)
		{
			if (kk[i] == 0 && flag == 0)
				continue;
			else
			{
				m1 = m1 + char(kk[i] + 48);
				flag = 1;
			}
		}
		return (m1 != "") ? m1 : "0";
	}
};