[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.

这道题是复数乘法，题目难度为Medium。

复数相信大家都不陌生，在乘法运算过程中i*i = -1，复数乘法可以表示为：

(a + bi) * (c + di)  = (ac - bd) + (ad + bc)i;

因此我们只需要从两个复数的字符串表示中分别分解出实部和虚部，然后按照公式生成最终结果即可。分解实部和虚部时，根据‘+’位置进行判断，‘+’之前的是实部，‘+’之后‘i’之前的是虚部。具体代码：

class Solution {
    void getAB(const string& s, int& a, int& b) {
        int i;
        for(i=0; i