537. Complex Number Multiplication

多多趣

于 2018-08-13 23:26:56 发布

阅读量85

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/excellentlizhensbfhw/article/details/81639049

LeetCode 专栏收录该内容

548 篇文章

订阅专栏

本文介绍了一种通过解析输入字符串并应用复数乘法规则来实现两个复数相乘的方法。给出了具体的Java代码示例，该方法能正确处理形如a+bi的复数输入，并返回同样形式的复数乘积。

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.

复数的运算（以前没注意复数的“复”的含义，现在明白了，原来是复杂的“复”），程序如下所示：

class Solution {
    public String complexNumberMultiply(String a, String b) {
        String[] arrayA = a.split("\\+");
        int aInt = Integer.parseInt(arrayA[0]);
        int aCom = Integer.parseInt(arrayA[1].substring(0, arrayA[1].length() - 1));
        String[] arrayB = b.split("\\+");
        int bInt = Integer.parseInt(arrayB[0]);
        int bCom = Integer.parseInt(arrayB[1].substring(0, arrayB[1].length() - 1));
        String resInt = String.valueOf(aInt*bInt - aCom*bCom);
        String resCom = String.valueOf(aInt*bCom + bInt*aCom);
        return resInt + "+" + resCom + "i";
    }
}