LeetCode 537. 复数乘法(Complex Number Multiplication)

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 + 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.

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

巴特西

LeetCode 537. 复数乘法(Complex Number Multiplication)

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