【leetcode】537. Complex Number Multiplication（Python & C++）

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537. Complex Number Multiplication

题目链接

537.1 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:

The input strings will not have extra blank.
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.

537.2 解题思路：

1. 思路一：分为两步，一步将字符串转化为数字，一步将数字进行复数运算，并返回结果字符串。

   • 获取数字函数getnumber：将字符串中加号之前的字符串转化为整型赋给real，将加号之后，i之前的字符串转化为整型赋给virtual。

   • 复数运算函数complexNumberMultiply：分别调用getnumber函数获取两个字符串对应的实数和虚数部分，然后进行复数运算，将结果转化为字符串并返回。

2. 思路二：分为c++和Python。

   • c++：主要是处理字符串与int型转换上有改进。利用了stringstream处理。

   • Python：利用了Python语言的特性，使用了map函数，并对字符串使用split函数进行分片，得到转换后的int型。

537.3 C++代码：

1、思路一代码（3ms）：

class Solution138 {
public:
    void getnumber(string s, int &real, int &ivirtual)
    {
        string sreal = "0";
        string svirtual = "0";
        for (int i = 0; i < s.length();i++)
        {
            if (s[i]=='+')
            {
                sreal = s.substr(0, i);
                svirtual = s.substr(i + 1, s.length() - i - 2);
            }
        }
        real = atoi(sreal.c_str());
        ivirtual = atoi(svirtual.c_str());
    }
    string complexNumberMultiply(string a, string b) {
        int a_real, a_ivirtual, b_real, b_ivirtual;
        getnumber(a, a_real, a_ivirtual);
        getnumber(b, b_real, b_ivirtual);
        int real = a_real*b_real - a_ivirtual*b_ivirtual;
        int ivirtual = a_ivirtual * b_real + b_ivirtual * a_real;
        string result = to_string(real) + "+" + to_string(ivirtual) + "i";
        return result;
    }
};

2、思路二代码（3ms）

class Solution138_1 {
public:
    string complexNumberMultiply(string a, string b) {
        int ra, ia, rb, ib;
        char buff;
        stringstream aa(a), bb(b), ans;
        aa >> ra >> buff >> ia >> buff;
        bb >> rb >> buff >> ib >> buff;
        ans << ra*rb - ia*ib << "+" << ia*rb + ib*ra << "i";
        return ans.str();
    }
};

537.4 Python代码：

1、思路一代码（32ms）

class Solution(object):
    def complexNumberMultiply(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: str
        """
        def getnumber(s):
            sreal="0"
            svirtual="0"
            for i in range(len(s)):
                if s[i]=='+':
                    sreal=s[0:i]
                    svirtual=s[i+1:len(s)-1]
            return int(sreal),int(svirtual)
        a_real,a_virtual=getnumber(a)
        b_real,b_virtual=getnumber(b)
        real=a_real*b_real-a_virtual*b_virtual
        virtual=a_virtual*b_real+b_virtual*a_real
        result=str(real)+"+"+str(virtual)+"i"
        return result

2、思路二代码（32ms）

class Solution1(object):
    def complexNumberMultiply(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: str
        """
        ra,ia=map(int,a[0:-1].split('+'))
        rb,ib=map(int,b[0:-1].split('+'))
        return str(ra*rb-ia*ib)+"+"+str(ia*rb+ib*ra)+"i"

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