LeetCode.537.Complex Number Multiplication
题目描述
- 537 Complex Number Multiplication
https://leetcode.com/problems/complex-number-multiplication/description/
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]. 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
- 537 Complex Number Multiplication
https://github.com/masikkk/leetcode-java/tree/master/problems/_537_ComplexNumberMultiplication
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