题目地址：https://leetcode.com/problems/complex-number-multiplication/description/
题目描述
Given two strings representing two complex numbers.
Youneed 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、 Theinputstringswillnothaveextrablank.；
2、 Theinputstringswillbegivenintheformofa+bi,wheretheintegeraandbwillbothbelongtotherangeof[-100,100].Andtheoutputshouldbealsointhisform.；
解题方法
方法一：
求两个复数的乘积。这个题中已经说明了一定会存在+号，这就是留个我们根据格式化的表达读取数字的实虚部用的。
另外，注意字符串的格式化时，后面所有的变量必须用括号括起来才行，也就是tuple型。
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)
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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';
}
};
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