作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/


题目地址:https://leetcode.com/problems/complex-number-multiplication/description/

题目描述

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:

  1. Input: "1+1i", "1+1i"
  2. Output: "0+2i"
  3. Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i.

Example 2:

  1. Input: "1+-1i", "1+-1i"
  2. Output: "0+-2i"
  3. 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.

解题方法

方法一:

求两个复数的乘积。这个题中已经说明了一定会存在+号,这就是留个我们根据格式化的表达读取数字的实虚部用的。

另外,注意字符串的格式化时,后面所有的变量必须用括号括起来才行,也就是tuple型。

  1. class Solution(object):
  2. def complexNumberMultiply(self, a, b):
  3. """
  4. :type a: str
  5. :type b: str
  6. :rtype: str
  7. """
  8. num1 = map(int, a[:-1].split('+'))
  9. num2 = map(int, b[:-1].split('+'))
  10. real = num1[0] * num2[0] - num1[1] * num2[1]
  11. virtual = num1[0] * num2[1] + num1[1] * num2[0]
  12. return "%d+%di" % (real, virtual)

C++处理的字符串的时候稍微麻烦了一些。

  1. class Solution {
  2. public:
  3. string complexNumberMultiply(string a, string b) {
  4. int ap = a.find('+');
  5. int ar = stoi(a.substr(0, ap));
  6. int ai = stoi(a.substr(ap + 1, a.size() - 1));
  7. int bp = b.find('+');
  8. int br = stoi(b.substr(0, bp));
  9. int bi = stoi(b.substr(bp + 1, b.size() - 1));
  10. string ansr = to_string(ar * br - ai * bi);
  11. string ansi = to_string(ar * bi + ai * br);
  12. return ansr + '+' + ansi + 'i';
  13. }
  14. };

日期

2018 年 2 月 5 日
2018 年 12 月 4 日 —— 周二啦!

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