<leetcode> Complex Number Multiplication

2018-03-30

因为这道题让我想起了编译里的词法分析，所以记下来吧。
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:

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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:

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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.
代码：

#include<sstream>
#include<string>
class Solution {
public:
    string complexNumberMultiply(string a, string b) {
        int a1,ai,b1,bi;
        int *pa=getnum(a);
        int *pb=getnum(b);
        a1=pa[0];ai=pa[1];
        b1=pb[0];bi=pb[1];
        int ans1=a1*b1-ai*bi;
        int ansi=a1*bi+ai*b1;
        stringstream ss;
        ss<<ans1;
        string str1;
        ss>>str1;
        ss.clear();
        ss<<ansi;
        string stri;
        ss>>stri;
        return str1+'+'+stri+'i';
    }
    int* getnum(string s){
        int s1,si;
        const char *p=s.data();
        int sf=0;
        int temp=0;
        bool flag1=false,flagi=false;
        while(*p!='\0'){
            if(*p=='-'&&sf==0){
                flag1=true;
            }
            if(*p=='-'&&sf!=0){
                flagi=true;
            }
            if(*p>='0'&&*p<='9'){
                temp=temp*10+(*p-48);
            }
            if(*p=='+'){
                s1=temp;
                temp=0;
            }
            if(*p=='i'){
                si=temp;
            }
            sf++;
            p++;
        }
        if(flag1){
            s1=-s1;
        }
        if(flagi){
            si=-si;
        }
        int *ans=new int[2];
        ans[0]=s1;
        ans[1]=si;
        return ans;
    }
};