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.

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给定两个复数字符串，要求计算它们的乘积。 简单题，展开之后是：(a1+a2i)(b1+b2i)=(a1*b1-a2*b2)+(a1*b2+a2*b1)i。所以只需要提取出a1,a2,b1,b2，然后代入公式计算到c1和c2，最后用to_string转换为字符串即可。 代码如下： [cpp] class Solution { void str2cpx(const string& s, int& c1, int&c2) { int i = 0, j = 0; while (s[j] != ‘+’)++j; c1 = atoi(s.substr(i, j – i).c_str()); i = ++j; while (s[j] != ‘i’)++j; c2 = atoi(s.substr(i, j – i).c_str()); } public: string complexNumberMultiply(string a, string b) { int a1 = 0, a2 = 0, b1 = 0, b2 = 0, c1 = 0, c2 = 0; str2cpx(a, a1, a2); str2cpx(b, b1, b2); c1 = a1*b1 – a2*b2; c2 = a1*b2 + a2*b1; return to_string(c1) + "+" + to_string(c2) + "i"; } }; [/cpp] 本代码提交AC，用时3MS。]]>