43 - Multiply Strings

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.

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思路分析：

乘法原理，感觉如果用数组来计算会简单很多，用字符串必须再每次循环后处理一下高位的进位问题，比较麻烦。

1、当个位数相乘时，digit无用。

即当123 x 54 时，当 4 x 123 digit 始终为0，此后 digit位表示每一位的进位值。

/*
*/

#include "stdafx.h"
#include <iostream>
#include <algorithm>

using namespace std;

class Solution_043_MultiplyStringsArray
{
public:
	string multiply(string num1, string num2) 
	{
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		string s(num1.size() + num2.size(), '0');

		reverse(num1.begin(), num1.end());
		reverse(num2.begin(), num2.end());

		for (int i = 0; i < num1.size(); i++)
		{
			int flag = 0;
			for (int j = 0; j < num2.size(); j++)
			{
				int digit = s[i + j] - '0';
				int num = (num1[i] - '0') * (num2[j] - '0');
				int res = digit + num + flag;
				s[i + j] = (res % 10) + '0';
				flag = res / 10;//进位
			}
			int index = i + num2.size();
			while (flag)
			{
				int digit = s[index] - '0';
				int res = digit + flag;
				s[index] = (res % 10) + '0';
				flag = res / 10;
			}
		}

		while (true)
		{
			if (s[s.size() - 1] == '0' && s.size() > 1)
				s.erase(s.size() - 1, 1);
			else
				break;
		}

		reverse(s.begin(), s.end());

		return s;
	}
};