LeetCode-537. Complex Number Multiplication-优快云博客

本文介绍了一种通过解析两个字符串形式的复数并计算它们乘积的方法。具体地，文章提供了一个示例代码，该代码将输入的复数字符串转换为整数，然后计算其乘积，并最终返回以字符串形式表示的复数乘积结果。

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]. Andthe output should be also in this form.

solution:

class Solution {
    public String complexNumberMultiply(String a, String b) {
        //x="a+bi", y="c+di"
        //x*y=(ac-bd)+(bc+ad)i
	String[] as=a.split("\\+|i");	
	int[] arri=new int[as.length];
	for(int i=0;i<as.length;i++)
	    arri[i]=Integer.parseInt(as[i]);
        //a=arri[0]; b=arri[1];
		
	String[] bs=b.split("\\+|i");	
	int[] arri2=new int[bs.length];
	for(int i=0;i<bs.length;i++)
            arri2[i]=Integer.parseInt(bs[i]);
        // c=arri2[0] d=arri2[1];
        
        return (arri[0]*arri2[0]-arri[1]*arri2[1])+"+"+(arri[1]*arri2[0]+arri[0]*arri2[1])+"i";
    }
}