LeetCode.537 Complex Number Multiplication

原创于 2017-12-02 22:43:40 发布 · 143 阅读

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本文介绍了一种通过解析字符串形式的复数并计算其乘积的方法。利用特定的字符串分割技术来提取实部和虚部，并根据复数乘法规则进行计算，最终返回标准化的复数乘积形式。

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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:

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.

分析：

class Solution {
    public String complexNumberMultiply(String a1, String b1) {
        //给定两个复数，求两者的乘积，最后返回结果。
        //(a+bi)(c+di)=ac+adi+bci-(bd)
        //注意：给定的字符串满足上述格式，a和b 范围均为-100、100
        int a=divideNum(a1)[0];
        int b=divideNum(a1)[1];
        int c=divideNum(b1)[0];
        int d=divideNum(b1)[1];
        
        int natureNum=a*c-b*d;
        int complexNum=a*d+b*c;
        return natureNum+"+"+complexNum+"i";

    }
    //切分数据
    public int[] divideNum(String a){
        int [] res=new int[2];
        //对于单个匹配字段，如+，*，|、\等，可以使用类似于“\\+"或者[+]进行处理，如图所示：
        String[] str=a.split("\\+");
        char [] chf=str[0].toCharArray();
        int former=0;
        if(chf[0]!='-'){
            //正数
            for(char c:chf){
                former=former*10+c-'0';
            }
        }else{
            //复数
            for(int i=1;i<chf.length;i++){
                former=former*10+chf[i]-'0';
            }
            former=-former;
        }
        
        String [] sl=str[1].split("i");
        int later=0;
        char [] chl=sl[0].toCharArray();
        if(chl[0]!='-'){
            //复数
            for(char c:chl){
                later=later*10+c-'0';
            }
        }else {
            //正数
            for(int i=1;i<chl.length;i++){
                later=later*10+chl[i]-'0';
            }
            later=-later;
        }
        res[0]=former;
        res[1]=later;
        return res;
        
    }
}