题目描述
A complex number can be represented as a string on the form “real+imaginaryi” where:
- real is the real part and is an integer in the range [-100, 100].
- imaginary is the imaginary part and is an integer in the range [-100, 100].
- i^2 == -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:
- num1 and num2 are valid complex numbers.
解题思路
- 字符串的基本操作，有find，substr，stoi（目前这个容易记，string2int）不回再记反什么的
C++代码 层次遍历
class Solution {
public:
pair<int, int> parse(string num){
int pos = num.find('+');
int real = stoi(num.substr(0, pos));
int imaginary = stoi(num.substr(pos + 1, num.size() - pos - 2));
pair<int, int> res(real, imaginary);
return res;
}
string complexNumberMultiply(string num1, string num2) {
pair<int, int> a = parse(num1), b = parse(num2);
int real_a = a.first, imaginary_a = a.second;
int real_b = b.first, imaginary_b = b.second;
return to_string(real_a * real_b - imaginary_a * imaginary_b) + "+" + to_string(real_a * imaginary_b + real_b * imaginary_a) + "i";
}
};
