[leetcode-537-Complex Number Multiplication]

Given two strings representing two complex numbers.

You need to return a string representing their multiplication. Note i2 = -1 according to the definition.

Example 1:

Input: "1+1i", "1+1i"
Output: "0+2i"
Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i.

Example 2:

Input: "1+-1i", "1+-1i"
Output: "0+-2i"
Explanation: (1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i, and you need convert it to the form of 0+-2i.

Note:

1. The input strings will not have extra blank.
2. The input strings will be given in the form of a+bi, where the integer a and b will both belong to the range of [-100, 100]. And the output should be also in this form.

思路：

首先将 ‘+’找到，将字符串分成左右两部分，然后分别转换成整数就可以了。

int stringToint(string a,int& real,int& virtu)
    {
        int i = 0;
        for (; i < a.size();i++)//找到+号 位置i
        {
            if (a[i] == '+') break;             
        }
        
        real = atoi(a.substr(0, i).c_str());
        virtu = atoi(a.substr(i+1,a.size()-i-2).c_str());
        //cout << real << " " << virtu << endl;
    
        return 0;
    }
    string complexNumberMultiply(string a, string b)
    {
        if (a.size() == 0 || b.size() == 0) return "0";
        int real1, virtu1, real2, virtu2,real,virtu;
        stringToint(a, real1, virtu1);
        stringToint(b, real2, virtu2);
        real = real1*real2 - virtu1*virtu2;
        virtu = real1*virtu2 + virtu1*real2;
        char resu[128];
        sprintf(resu, "%d+%di", real, virtu);
        return resu;
    }