leetcode537. Complex Number Multiplication

537. Complex Number Multiplication

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.

解法

先解析，得到实部和虚部，然后分别计算实部和虚部。

public class Solution {
    public String complexNumberMultiply(String a, String b) {
        if (a == null || a.length() == 0) {
            return null;
        }
        if (b == null || b.length() == 0) {
            return null;
        }

        int[] n1 = revert(a);
        int[] n2 = revert(b);
        int shi = n1[0] * n2[0] - n1[1] * n2[1];
        int xu = n1[0] * n2[1] + n1[1] * n2[0];
        String ret = shi + "+" + xu + "i";

        return ret;
    }
    private int[] revert(String a) {
        String[] ab = a.split("\\+");
        String[] bs = ab[1].split("i");
        int[] ret = new int[2];
        ret[0] = Integer.parseInt(ab[0]);
        ret[1] = Integer.parseInt(bs[0]);

        return ret;
    }
}