537 复数乘法（中等

解题思路：用加号作为分隔符，然后取出前后的所有数使用复数运算法则进行计算即可。题解如下，官方代码更简洁。

#
# @lc app=leetcode.cn id=537 lang=python3
#
# [537] 复数乘法
#

# @lc code=start
from re import I


class Solution:
    def complexNumberMultiply(self, num1: str, num2: str) -> str:
        #我的代码
        # num1 = num1.split('+')
        # num2 = num2.split('+')
        # a1, a2 = int(num1[0]), int(num2[0])
        # b1, b2 = int(num1[1][:(len(num1[1])-1)]
        #              ), int(num2[1][:(len(num2[1])-1)])
        # a = a1*a2-b1*b2
        # b = a1*b2+a2*b1
        # return str(a)+'+'+str(b)+'i'
        # leetcode官方题解
        real1, img1 = map(int, num1[:-1].split('+'))
        real2, img2 = map(int, num2[:-1].split('+'))
        return f'{real1*real2-img1*img2}+{real1*img2+img1*real2}i'

# @lc code=end

#学习python代码中map函数的用法

//官方C++代码，使用正则表达式
class Solution {
public:
    string complexNumberMultiply(string num1, string num2) {
        regex re("\\+|i"); 
        vector<string> complex1(sregex_token_iterator(num1.begin(), num1.end(), re, -1), std::sregex_token_iterator());
        vector<string> complex2(sregex_token_iterator(num2.begin(), num2.end(), re, -1), std::sregex_token_iterator());
        int real1 = stoi(complex1[0]);
        int imag1 = stoi(complex1[1]);
        int real2 = stoi(complex2[0]);
        int imag2 = stoi(complex2[1]);
        return to_string(real1 * real2 - imag1 * imag2) + "+" + to_string(real1 * imag2 + imag1 * real2) + "i";
    }
};

作者：LeetCode-Solution
链接：https://leetcode.cn/problems/complex-number-multiplication/solution/fu-shu-cheng-fa-by-leetcode-solution-163i/
来源：力扣（LeetCode）
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