[LeetCode] Complex Number Multiplication

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.

利用stringstream格式化字符串后计算。

class Solution {
public:
    string complexNumberMultiply(string a, string b) {
        int ra, ia, rb, ib;
        char sign;
        stringstream sa(a), sb(b), res;
        sa >> ra >> sign >> ia >> sign;
        sb >> rb >> sign >> ib >> sign;
        res << ra * rb - ia * ib << "+" << ra * ib + rb * ia << "i";
        return res.str();
    }
};
// 3 ms

 利用sscanf格式化字符串后计算。

class Solution {
public:
    string complexNumberMultiply(string a, string b) {
        int ra, ia, rb, ib;
        sscanf(a.c_str(), "%d+%di", &ra, &ia);
        sscanf(b.c_str(), "%d+%di", &rb, &ib);
        return to_string(ra * rb - ia * ib) + "+" + to_string(ra * ib + rb * ia) + "i";
    }
};
// 3 ms

 使用stoi来转换

使用substr和find分割字符串

class Solution {
public:
    string complexNumberMultiply(string a, string b) {
        auto pos = a.find('+');
        int ra = stoi(a.substr(0, pos));
        int ia = stoi(a.substr(pos + 1, a.find('i')));
        pos = b.find('+');
        int rb = stoi(b.substr(0, pos));
        int ib = stoi(b.substr(pos + 1, b.find('i')));
        return to_string(ra * rb - ia * ib) + "+" + to_string(ra * ib + rb * ia) + "i";
    }
};
// 3 ms