537. Complex Number Multiplication

最新推荐文章于 2020-01-26 19:38:39 发布

原创最新推荐文章于 2020-01-26 19:38:39 发布 · 990 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LeetCode #Math #String #Complex Number Multi

LeetCode 同时被 3 个专栏收录

402 篇文章

订阅专栏

Math

50 篇文章

订阅专栏

String

36 篇文章

订阅专栏

本文详细介绍了如何通过编程实现两个复数的乘法运算，并给出了具体的代码实现。通过实例演示了不同复数相乘后的结果转换形式。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

复数乘法公式：(a1+b1i)*(a2+b2i) = (a1a2 - b1b2) + (a1b2+a2b1)i

按照公式写代码如下：

public class Solution {
    public String complexNumberMultiply(String a, String b) {
        String[] A = a.split("\\+");
        String[] B = b.split("\\+");
        int a1 = Integer.parseInt(A[0]);
        int b1 = Integer.parseInt(A[1].replace("i",""));
        int a2 = Integer.parseInt(B[0]);
        int b2 = Integer.parseInt(B[1].replace("i",""));
        String res = (a1*a2 - b1*b2) + "+" + (a1*b2 + a2*b1) + "i";
        return res;
    }
}