Algorithm
Problem Name: 537. Complex Number Multiplication
A complex number can be represented as a string on the form "real+imaginaryi" where:
realis the real part and is an integer in the range[-100, 100].imaginaryis the imaginary part and is an integer in the range[-100, 100].i2 == -1.
Given two complex numbers num1 and num2 as strings, return a string of the complex number that represents their multiplications.
Example 1:
Input: num1 = "1+1i", num2 = "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: num1 = "1+-1i", num2 = "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.
Constraints:
num1andnum2are valid complex numbers.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public String complexNumberMultiply(String num1, String num2) {
ComplexNumber result = ComplexNumber.buildComplexNumber(num1).multiply(ComplexNumber.buildComplexNumber(num2));
return result.toString();
}
private static class ComplexNumber {
private final int real;
private final int imaginary;
private ComplexNumber(int real, int imaginary) {
this.real = real;
this.imaginary = imaginary;
}
public static ComplexNumber buildComplexNumber(String s) {
int[] idx = {0};
int realPart = parseRealPart(s, idx);
int imaginaryPart = parseImaginaryPart(s, idx);
return new ComplexNumber(realPart, imaginaryPart);
}
public ComplexNumber multiply(ComplexNumber anotherNumber) {
int realPart = this.real * anotherNumber.real + (this.imaginary * anotherNumber.imaginary * -1);
int imaginaryPart = this.real * anotherNumber.imaginary + this.imaginary * anotherNumber.real;
return new ComplexNumber(realPart, imaginaryPart);
}
public String toString() {
return this.real + "+" + this.imaginary + "i";
}
private static int parseRealPart(String s, int[] idx) {
int realSign = 1;
if (s.charAt(idx[0]) == '+' || s.charAt(idx[0]) == '-') {
realSign = s.charAt(idx[0]) == '-' ? -1 : 1;
idx[0]++;
}
int realNum = 0;
while (idx[0] < s.length() && Character.isDigit(s.charAt(idx[0]))) {
realNum = realNum * 10 + Character.getNumericValue(s.charAt(idx[0]++));
}
realNum *= realSign;
idx[0]++;
return realNum;
}
private static int parseImaginaryPart(String s, int[] idx) {
int imaginarySign = 1;
if (idx[0] < s.length() && (s.charAt(idx[0]) == '+' || s.charAt(idx[0]) == '-')) {
imaginarySign = s.charAt(idx[0]) == '-' ? -1 : 1;
idx[0]++;
}
int imaginaryNum = 0;
while (idx[0] < s.length() && s.charAt(idx[0]) != 'i') {
imaginaryNum = imaginaryNum * 10 + Character.getNumericValue(s.charAt(idx[0]++));
}
return imaginarySign * imaginaryNum;
}
}
}
Input
num1 = "1+1i", num2 = "1+1i"
Output
"0+2i"
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def complexNumberMultiply(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
re, im = 0, 0
re_a, im_a = list(map(int,a[:-1].split("+")))
re_b, im_b = list(map(int,b[:-1].split("+")))
re += re_a * re_b - im_a * im_b
im += re_a * im_b + re_b *im_a
return str(re)+"+"+str(im)+"i"
Input
num1 = "1+1i", num2 = "1+1i"
Output
"0+2i"
#3 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0537_ComplexNumberMultiplication
{
public string ComplexNumberMultiply(string a, string b)
{
var split1 = a.Split(new char[] { '+', 'i' }, StringSplitOptions.RemoveEmptyEntries);
var split2 = b.Split(new char[] { '+', 'i' }, StringSplitOptions.RemoveEmptyEntries);
int a1 = int.Parse(split1[0]), b1 = int.Parse(split1[1]);
int a2 = int.Parse(split2[0]), b2 = int.Parse(split2[1]);
return $"{a1 * a2 - b1 * b2}+{a1 * b2 + a2 * b1}i";
}
}
}
Input
num1 = "1+-1i", num2 = "1+-1i"
Output
"0+-2i"