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LeetCode.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.

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.

解题过程

很简单的字符串分割和数学题,也没啥需要注意的边界点,用java写起来也是真方便,一次AC。

package _537_ComplexNumberMultiplication;

class Solution {
    //a=a1+a2i, b=b1+b2i,结果c=c1+c2i
    public String complexNumberMultiply(String a, String b) {
        int a1 = Integer.parseInt(a.substring(0, a.indexOf("+")));
        int a2 = Integer.parseInt(a.substring(a.indexOf("+")+1, a.length()-1));
        int b1 = Integer.parseInt(b.substring(0, b.indexOf("+")));
        int b2 = Integer.parseInt(b.substring(b.indexOf("+")+1, b.length()-1));        
        int c1 = a1*b1 - a2*b2;
        int c2 = a1*b2 + b1*a2;
        return new String(c1+"+"+c2+"i");
    }
}

public class ComplexNumberMultiplication {
    public static void main(String[] args){
        Solution solution = new Solution();
        System.out.println(solution.complexNumberMultiply("1+1i", "1+1i"));
        System.out.println(solution.complexNumberMultiply("1+-1i", "1+-1i"));
    }
}

GitHub代码


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创建日期 2018-01-09
修改日期 2018-01-11
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