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:

The input strings will not have extra blank.
The input strings will be given in the form of a+bi, where the integera and
b will both belong to the range of [-100, 100]. Andthe output should be also in this form.
思路：提取出实部与虚部

知识点：1.字符转整形；2.在string中提取整数

代码1：

class Solution {
public:
string complexNumberMultiply(string a, string b) {
int i=0,len=0,a1,a2,b1,b2;
string sub;
i = a.find("+");
len = a.size();
a1 = stoi(a.substr(0,i));
a2 = stoi(a.substr(i+1,len-1));
i = b.find("+");
len = b.size();
b1 = stoi(b.substr(0,i));
b2 = stoi(b.substr(i+1,len-1));
return (to_string(a1*b1-a2*b2)+"+"+to_string(a1*b2+a2*b1)+"i");

}
};

string.find; string.substring; stoi（此处有问题）; to_string
代码2：

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();
}
};

核心在于如何利用stringstream从字符串中提取数字。 http://www.cplusplus.com/reference/sstream/stringstream/ http://blog.csdn.net/sophia1224/article/details/53054698 https://www.cnblogs.com/hujunzheng/p/5042068.html