LeetCode 537. 复数乘法题解

题目来源：537. 复数乘法

2022.02.25 每日一题

每日一题专栏地址：LeetCode 每日一题题解更新中❤️

今天就要返校了，一会更新完题解就走了，回去就见到我的好大儿们了，哈哈哈

今天的题目就是两个复数相乘

复数1： A = a + b i A=a+bi A=a+bi，复数2： B = c + d i B=c+di B=c+di

可得 A B = ( a c − b d ) + ( a d + b c ) i AB = (ac-bd)+(ad+bc)i AB=(ac−bd)+(ad+bc)i

那么我们只需要找到对于的 a、b、c、d即可

class Solution {
public:
    string complexNumberMultiply(string num1, string num2) {
        int len1 = num1.size(), len2 = num2.size();
        int i;
        for (i = 0; i < len1; i++)
            if (num1[i] == '+')
                break;
        int a = stoi(num1.substr(0, i));
        int b = stoi(num1.substr(i + 1, len1 - 1));
        for (i = 0; i < len2; i++)
            if (num2[i] == '+')
                break;
        int c = stoi(num2.substr(0, i));
        int d = stoi(num2.substr(i + 1, len2 - 1));
        // (ac-bd)+(ad+bc)i
        return to_string(a * c - b * d) + "+" + to_string(a * d + b * c) + "i";
    }
};

class Solution {
    public String complexNumberMultiply(String num1, String num2) {
        int len1 = num1.length(), len2 = num2.length();
        int i;
        for (i = 0; i < len1; i++)
            if (num1.charAt(i) == '+') break;
        int a = Integer.parseInt(num1.substring(0, i));
        int b = Integer.parseInt(num1.substring(i + 1, len1 - 1));
        for (i = 0; i < len2; i++)
            if (num2.charAt(i) == '+') break;
        int c = Integer.parseInt(num2.substring(0, i));
        int d = Integer.parseInt(num2.substring(i + 1, len2 - 1));
        // (ac-bd)+(ad+bc)i
        return (a * c - b * d) + "+" + (a * d + b * c) + "i";
    }
}