# 537. Complex Number Multiplication 复数乘法

@TOC

## # 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.

## # 解题方法

``````class Solution(object):
def complexNumberMultiply(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
num1 = map(int, a[:-1].split('+'))
num2 = map(int, b[:-1].split('+'))
real = num1[0] * num2[0] - num1[1] * num2[1]
virtual = num1[0] * num2[1] + num1[1] * num2[0]
return "%d+%di" % (real, virtual)
``````

C++处理的字符串的时候稍微麻烦了一些。

``````class Solution {
public:
string complexNumberMultiply(string a, string b) {
int ap = a.find('+');
int ar = stoi(a.substr(0, ap));
int ai = stoi(a.substr(ap + 1, a.size() - 1));
int bp = b.find('+');
int br = stoi(b.substr(0, bp));
int bi = stoi(b.substr(bp + 1, b.size() - 1));
string ansr = to_string(ar * br - ai * bi);
string ansi = to_string(ar * bi + ai * br);
return ansr + '+' + ansi + 'i';
}
};
``````

## # 日期

2018 年 2 月 5 日 2018 年 12 月 4 日 —— 周二啦！