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 + i
2
+ 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 + i
2
- 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.
这道题让我们求复数的乘法,有关复数的知识最早还是在本科的复变函数中接触到的,难起来还真是难。但是这里只是最简单的乘法,只要利用好定义i2=-1就可以解题,而且这道题的另一个考察点其实是对字符的处理,我们需要把字符串中的实部和虚部分离开并进行运算,那么我们可以用STL中自带的find_last_of函数来找到加号的位置,然后分别拆出实部虚部,进行运算后再变回字符串,参见代码如下:
解法一:
public: string complexNumberMultiply(string a, string b) { int n1 = a.size(), n2 = b.size(); auto p1 = a.find_last_of("+"), p2 = b.find_last_of("+"); int a1 = stoi(a.substr(0, p1)), b1 = stoi(b.substr(0, p2)); int a2 = stoi(a.substr(p1 + 1, n1 - p1 - 2)); int b2 = stoi(b.substr(p2 + 1, n2 - p2 - 2)); int r1 = a1 * b1 - a2 * b2, r2 = a1 * b2 + a2 * b1; return to_string(r1) + "+" + to_string(r2) + "i"; } };
下面这种方法利用到了字符串流类istringstream来读入字符串,直接将实部虚部读入int变量中,注意中间也要把加号读入char变量中,然后再进行运算即可,参见代码如下
解法二:
public: string complexNumberMultiply(string a, string b) { istringstream is1(a), is2(b); int a1, a2, b1, b2, r1, r2; char plus; is1 >> a1 >> plus >> a2; is2 >> b1 >> plus >> b2; r1 = a1 * b1 - a2 * b2, r2 = a1 * b2 + a2 * b1; return to_string(r1) + "+" + to_string(r2) + "i"; } };
下面这种解法实际上是C语言的解法,用到了sscanf这个读入字符串的函数,需要把string转为cost char*型,然后标明读入的方式和类型,再进行运算即可,参见代码如下:
解法三:
public: string complexNumberMultiply(string a, string b) { int a1, a2, b1, b2, r1, r2; sscanf(a.c_str(), "%d+%di", &a1, &a2); sscanf(b.c_str(), "%d+%di", &b1, &b2); r1 = a1 * b1 - a2 * b2, r2 = a1 * b2 + a2 * b1; return to_string(r1) + "+" + to_string(r2) + "i"; } };
参考资料:
https://discuss.leetcode.com/topic/84261/java-3-liner
https://discuss.leetcode.com/topic/84382/c-using-stringstream
https://discuss.leetcode.com/topic/84323/java-elegant-solution
https://discuss.leetcode.com/topic/84508/cpp-solution-with-sscanf
本文转自博客园Grandyang,原文链接:[LeetCode] Complex Number Multiplication 复数相乘
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