537. 复数乘法

537. Complex Number Multiplication

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.

LeetCode537. Complex Number Multiplication中等

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:

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.

Java 实现

class Solution {    // 复数的乘法: (a+bi)(c+di)=(ac-bd)+(bc+ad)i    public String complexNumberMultiply(String a, String b) {        int[] arrA = getValue(a);        int[] arrB = getValue(b);        int x = arrA[0] * arrB[0] - arrA[1] * arrB[1];        int y = arrA[1] * arrB[0] + arrA[0] * arrB[1];        return x + "+" + y + "i";    }    private int[] getValue(String s) {        String[] str = s.split("\\+");        int[] val = new int[2];        val[0] = Integer.parseInt(str[0]);        val[1] = Integer.parseInt(str[1].replace("i", ""));        return val;    }}

参考资料