537. Complex Number Multiplication

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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:

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.

题目含义：求两个复数的乘法

思路：根据复数乘法公式  (a+bi)(c+di) = (ac-bd)+(ad+bc)i

 1     public String complexNumberMultiply(String a, String b) {
 2         String[] A = a.split("\\+");
 3         String[] B = b.split("\\+");
 4         int a1 = Integer.parseInt(A[0]);
 5         int b1 = Integer.parseInt(A[1].replace("i",""));
 6 
 7         int a2 = Integer.parseInt(B[0]);
 8         int b2 = Integer.parseInt(B[1].replace("i",""));
 9         return (a1*a2 - b1*b2)+"+"+(a1*b2 + a2*b1) +"i";        
10         
11     }

537. Complex Number Multiplication

标签：this   extra   you   style   not   乘法   公式   for   input   

原文地址：http://www.cnblogs.com/wzj4858/p/7686305.html