[Leetcode] 537. Complex Number Multiplication 解题报告_leetcode 537 fuxuemingzhu-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/78851103

本文介绍了一种利用C++的stringstream简化复数乘法运算的方法。通过解析输入字符串获取实部与虚部，并计算结果，最终以字符串形式返回。此方法简洁高效，易于理解和实现。

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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:

The input strings will not have extra blank.
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.

思路：

哈哈，easy level的题目，思路上无需解释了。但是采用stringstream之后，代码变得出奇的简洁，请参考。

代码：

class Solution {
public:
    string complexNumberMultiply(string a, string b) {
        int r1, i1, r2, i2, r, i;
        getComplexNumber(a, r1, i1);
        getComplexNumber(b, r2, i2);
        r = r1 * r2 - i1 * i2;
        i = r1 * i2 + r2 * i1;
        return to_string(r) + '+' + to_string(i) + 'i';
    }
private:
    void getComplexNumber(const string &s, int &r, int &i) {
        r = 0, i = 0;
        int index = 0;
        int real_sign = 1, imag_sign = 1;
        if (s[index] == '-') {
            real_sign = -1, ++index;
        }
        while (index < s.length() && isdigit(s[index])) {
            r = 10 * r + s[index++] - '0';
        }
        ++index;        // skip '+';
        if (s[index] == '-') {
            imag_sign = -1, ++index;
        }
        while (index < s.length() && isdigit(s[index])) {
            i = 10 * i + s[index++] - '0';
        }
        r *= real_sign, i *= imag_sign;
        return;
    }
};

采用stringstream版本：

class Solution {
public:
    string complexNumberMultiply(string a, string b) {
        int ra, ia, rb, ib;
        char buff;
        stringstream aa(a), bb(b), ans;
        aa >> ra >> buff >> ia >> buff;
        bb >> rb >> buff >> ib >> buff;
        ans << ra * rb - ia * ib << "+" << ra * ib + rb * ia << "i";
        return ans.str();
    }
};