LeetCoode 537. Complex Number Multiplication

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.

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:

1、The input strings will not have extra blank.
2、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.

题意

这道题就是做复数的乘法，首先获取到每个复数的实部和虚部，然后按照公式进行运算即可。

（a+bi）·（c+di）=（ac-bd）+（bc+ad）i

代码

string complexNumberMultiply(string a, string b) {    int adiv = a.find('+'), bdiv = b.find('+');    int aInt = atoi(a.substr(0,adiv).c_str()), bInt = atoi(b.substr(0,bdiv).c_str()),    aImg = atoi(a.substr(adiv+1).c_str()), bImg = atoi(b.substr(bdiv+1).c_str());    return to_string(aInt*bInt - aImg*bImg)+"+"+to_string(aInt*bImg+aImg*bInt)+"i";}

代码耗时
55 / 55 test cases passed.
Runtime: 3 ms