537. 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.
   
   

题目的要求很简单，就是给定两个复数（字符串形式string），求两个复数的乘积，并按照题目所要求的格式返回结果（字符串形式）。

其实解题思路很简单，就是解析每个复数的字符串，求出实部跟虚部，就可以计算得到最终结果。在这里，我定义了一个结构体来存储每个复数的实部跟虚部：

struct complexNumber {        int x;   // 实部        int y;   // 虚部    };

我利用了vector来实现栈的结构，以便解析出字符串中的实部和虚部。

遍历字符串中的每一个字符，只要不是+或者i，若是负号-，则-1入栈，若是数字字符，则转换为int类型入栈；

如果是+／i，说明已经可以开始计算实部／虚部，遍历栈，计算实部／虚部，计算部分的代码：

complexNumber getNumber(string s) {        complexNumber number;        number.x = 0;        number.y = 0;        vector<int> temp;        for (int i = 0; i < s.size(); i++) {            if (s[i] ==  '+') {                int count = 0;                while (!temp.empty()) {                    if (temp.back() == -1) {                        number.x *= -1;                    } else {                        number.x += temp.back() * pow(10, count);                        count += 1;                    }                    temp.pop_back();                }            } else if (s[i] == 'i') {                int count = 0;                while (!temp.empty()) {                    if (temp.back() == -1) {                        number.y *= -1;                    } else {                        number.y += temp.back() * pow(10, count);                        count += 1;                    }                    temp.pop_back();                }            } else {                if (s[i] == '-')                    temp.push_back(-1);                else                    temp.push_back(s[i] - '0');            }        }        return number;    }

最后完整的代码：

class Solution {public:    struct complexNumber {        int x;        int y;    };        complexNumber getNumber(string s) {        complexNumber number;        number.x = 0;        number.y = 0;        vector<int> temp;        for (int i = 0; i < s.size(); i++) {            if (s[i] ==  '+') {                int count = 0;                while (!temp.empty()) {                    if (temp.back() == -1) {                        number.x *= -1;                    } else {                        number.x += temp.back() * pow(10, count);                        count += 1;                    }                    temp.pop_back();                }            } else if (s[i] == 'i') {                int count = 0;                while (!temp.empty()) {                    if (temp.back() == -1) {                        number.y *= -1;                    } else {                        number.y += temp.back() * pow(10, count);                        count += 1;                    }                    temp.pop_back();                }            } else {                if (s[i] == '-')                    temp.push_back(-1);                else                    temp.push_back(s[i] - '0');            }        }        return number;    }        string complexNumberMultiply(string a, string b) {        int re1, im1;   // string a        int re2, im2;   // string b        complexNumber aNumber = getNumber(a);        complexNumber bNumber = getNumber(b);                re1 = aNumber.x;        im1 = aNumber.y;        re2 = bNumber.x;        im2 = bNumber.y;                // for result        int re = re1 * re2 - im1 * im2;        int im = im1 * re2 + re1 * im2;                stringstream result;        result << re << "+" << im << "i";                return result.str();    }};