537. Complex Number Multiplication

题目描述：

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.

解题思路：

题目本质很简单，解析输入参数稍微繁琐点。

代码：

class Solution {
public:
    string complexNumberMultiply(string a, string b) {
        int num1, num2, num3, num4;
        int res1, res2;
        string sig1, sig2, res;
        parse(num1, num2, sig1, a);
        parse(num3, num4, sig2, b);
        res1 = num1 * num3 - num2 * num4;
        res2 = num1 * num4 + num2 * num3;
        res = to_string(res1) + "+" +to_string(res2) + "i";
        return res;
    }
    void parse(int& a, int& b, string& sig, const string& input) {
        size_t plus =  input.find("+");
        size_t i = input.find("i");
        size_t minus = input.find("-");
        a = stoi(input.substr(0, plus));
        if (minus != string::npos) {
            b = stoi(input.substr(plus+1, i-(plus+1)));
            sig = "-";
        } else {
            b = stoi(input.substr(plus+1, i-(plus+1)));
            sig = "+";
        }
    }
};