[leetcode]: 537. Complex Number Multiplication

比眉伴天荒 2022-06-14 12:19 119阅读 0赞

# 1.题目 #

Given two strings representing two complex numbers.

You need to return a string representing their multiplication. Note i2 = -1 according to the definition.  
给出字符串形式的复数，求两个复数做乘法的结果，以字符串形式返回。

> 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.

# 2.分析 #

本题主要考点： 从字符串中提取复数  
比较笨的方法是遍历字符串，手动提取实部，虚部  
便利的方法例如：  
c++ stringstream  
python split

# 3.代码 #

c++ stringstream

class Solution {
    public:
        string complexNumberMultiply(string a, string b) {
            int a_Re, a_Im;
            int b_Re, b_Im;
            char buff;
            stringstream A(a), B(b),ans;
            A >> a_Re >> buff >> a_Im >> buff;
            B >> b_Re >> buff >> b_Im >> buff;
            ans << a_Re*b_Re - a_Im*b_Im << "+" << a_Re*b_Im + a_Im*b_Re << "i";
            return ans.str();
        }
    };

python split

class Solution(object):
        def complexNumberMultiply(self, a, b):
            ra,ia=map(int,a[:-1].split('+'))
            rb,ib=map(int,b[:-1].split('+'))
            return '%d+%di'%(ra*rb-ia*ib,ra*ib+rb*ia)

手动遍历

class Solution {
    public:
        void get_complex(string s, int& Re, int& Im) {
            int i = 0;
            int sign_Re = 1;
            int sign_Im = 1;
            if (s[i] == '-') {
                sign_Re = -1;
                ++i;
            }
            while (s[i] != '+') {
                Re = Re * 10 + s[i] - '0';
                ++i;
            }
            ++i;
            if (s[i] == '-')
            {
                sign_Im = -1;
                ++i;
            }
            while (s[i] != 'i') {
                Im = Im * 10 + s[i] - '0';
                ++i;
            }
            Re *= sign_Re;
            Im *= sign_Im;
        }
        string complexNumberMultiply(string a, string b) {
            int a_Re=0, a_Im = 0;
            int b_Re=0, b_Im = 0;
            get_complex(a, a_Re, a_Im);
            get_complex(b, b_Re, b_Im);
            int Re = a_Re*b_Re - a_Im*b_Im;
            int Im = a_Re*b_Im + a_Im*b_Re;
            return to_string(Re) + "+" + to_string(Im) + "i";
        }
    };