leetcode 537. Complex Number Multiplication 复数乘法 + C++中stringstream很好示例

矫情吗；* 2022-06-03 03:39 188阅读 0赞

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.

本题意很简单，就是计算复数的乘法，

这道题最大的启发就是C++的stringstream真的很好用

建议和这一道题[leetcode 539. Minimum Time Difference C++中的stringstream真的很好用][leetcode 539. Minimum Time Difference C_stringstream] 一起学习

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    
    using namespace std;
    
    
    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();
        }
    };

[leetcode 539. Minimum Time Difference C_stringstream]: http://blog.csdn.net/jackzhang_123/article/details/78825210