LeetCode Complex Number Multiplication

原题链接在这里:https://leetcode.com/problems/complex-number-multiplication/description/

题目:

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.

题解:

(aReal + aImag*i)*(bReal + bImag*i) = (aReal*bReal- aImag*bImag) + (aReal*bImag+bReal*aImag)*i.

从原有string里提取出实数和虚数两部分组成答案.

Time Complexity: O(a.length() + b.length()). split用时

Space: O(a.length() + b.length()). 中间的string arr.

AC Java:

 1 class Solution {
 2     public String complexNumberMultiply(String a, String b) {
 3         String [] aArr = a.split("\+|i");
 4         String [] bArr = b.split("\+|i");
 5         int aReal = Integer.valueOf(aArr[0]);
 6         int aImag = Integer.valueOf(aArr[1]);
 7         int bReal = Integer.valueOf(bArr[0]);
 8         int bImag = Integer.valueOf(bArr[1]);
 9         return (aReal*bReal-aImag*bImag) + "+" + (aReal*bImag+aImag*bReal) + "i";
10     }
11 }

  

原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/7723113.html