python--leetcode 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:

1. The input strings will not have extra blank.

1. 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.

啊哈，这一题的意思就是给两个字符串是两个复数的形式，求他们的乘积。返回一个字符串。

medium的题果然比easy难了一丢丢，不过这一题想来复杂，要实现其实不难。

代码：

class Solution(object):    def complexNumberMultiply(self, a, b):        """        :type a: str        :type b: str        :rtype: str        """        aa=a.split("+")        bb=b.split("+")        aa[1]=aa[1][:-1]        bb[1]=bb[1][:-1]        a1,b1,a2,b2=int(aa[0]),int(aa[1]),int(bb[0]),int(bb[1])        print(a1,a2,b1,b2)        part1=a1*a2-b1*b2        part2=a1*b2+b1*a2        return str(part1)+'+'+str(part2)+'i's=Solution()print(s.complexNumberMultiply("1+0i", "1+0i"))

思路非常的简单，就是对字符串进行处理，得到数字。

然后你可以在纸上推演一下这个公式，两个复数形式的复数相乘的结果。

再上一个更简单的代码：

class Solution(object):    def complexNumberMultiply(self, a, b):        """        :type a: str        :type b: str        :rtype: str        """        a1, a2 = map(int, a[:-1].split('+'))        b1, b2 = map(int, b[:-1].split('+'))        return '%d+%di' % (a1 * b1 - a2 * b2, a1 * b2 + a2 * b1)