[leetcode] 573 complex multiplication (conversion between characters and numbers)

537. Complex multiplication

The plural Can be expressed as a string , follow Real component + Imaginary part i" In the form of , And meet the following conditions ：

• Real component It's an integer , The value range is [-100, 100]
• Imaginary part It's also an integer , The value range is [-100, 100]
• i² == -1
• Give you a complex number represented by two strings num1 and num2, Please follow the plural form , Returns a string representing their product .

Example 1：

 Input ：num1 = "1+1i", num2 = "1+1i"
 Output ："0+2i"
 explain ：(1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i , You need to convert it to  0+2i  In the form of .

Example 2：

 Input ：num1 = "1+-1i", num2 = "1+-1i"
 Output ："0+-2i"
 explain ：(1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i , You need to convert it to  0+-2i  In the form of . 

Tips ：

• num1 and num2 Are valid plural representations .

Topic analysis ：

The first time I saw the title , Thought it would be troublesome , But the string interception function is not well remembered , The function of converting characters into numbers has also been forgotten , So when I came and went, I thought of violence , All implementations are implemented in code , Instead of calling the function , Cause the code to stink and long

The idea of the topic is very simple , It's multiplication of plural numbers , It can be regarded as two one variable first-order equations , Multiply the corresponding coefficient , Then we get the final expression , So the trouble is how to extract numbers from a string of characters

Method 1 ： violence

class Solution {
    
public:
    string complexNumberMultiply(string num1, string num2) {
    
        int n=num1.size(),m=num2.size();
        int a=0,b=0,c=0,d=0;// Find four numbers 
        int flag1=0,flag2=0;// Mark the , It is convenient to find two numbers of the same string 
        int ans1=1,ans2=1;// The sign , See if it's a negative number or a positive number 
        int cnt1=0,cnt2=0;
        char sss[100000];
        for(int i=0;i<n;++i){
    
            if(num1[i]>='0'&&num1[i]<='9'&&flag1==0){
    // Find the first number 
                a=a*10+(num1[i]-'0');
            }
            else{
    
                if(i!=0){
    // Find the second number 
                if(num1[i]=='-'||num1[i]=='+'&&num1[i+1]=='-')// Determine whether the second number is positive or negative 
                ans1=-1;
                flag1=1;
                 if(num1[i]>='0'&&num1[i]<='9'){
    // Find the second number 
                b=b*10+(num1[i]-'0');
                } 
            }
         }
        }
        if(num1[0]=='-')// If the first character is -, So it's plural 
        a=-a;
        b*=ans1;// Here is to see ans1 The state of , If there is no negative sign , that ans1 It won't change , conversely 
            for(int i=0;i<m;++i){
    
            if(num2[i]>='0'&&num2[i]<='9'&&flag2==0){
    
                c=c*10+(num2[i]-'0');
            }
            else{
    
                if(i!=0){
    
                    if(num2[i]=='-'||num2[i]=='+'&&num2[i+1]=='-')
                ans2=-1;
                flag2=1;
                 if(num2[i]>='0'&&num2[i]<='9'){
    
                d=d*10+(num2[i]-'0');
                }  
            }
            }
        }
        if(num2[0]=='-')
        c=-c;
        d*=ans2;
        cnt1=a*c-b*d;
        cnt2=a*d+c*b;
    return to_string(cnt1)+"+"+to_string(cnt2)+"i";
    }
};

Method 2 ： Call function

Calling a function makes the code look simple , High readability

find（）： Find the position where the current character first appears in the string

stoi（）： Convert a string into a number

substr（）： This function has two arguments , The first parameter represents the starting position of the intercepted string , The second parameter represents the length of the string to be intercepted

The second number is from pos1+1 Began to intercept , And the intercept length is pos1+1 To n-1（n-1 Is the last character of the string ）, So the length is n-pos1-2

class Solution {
    
public:
    string complexNumberMultiply(string num1, string num2) {
    
        int n=num1.size(),m=num2.size();
        int pos1=num1.find("+");
        int pos2=num2.find("+");
        int a = stoi(num1.substr(0,pos1)),b=stoi(num1.substr(pos1+1,n-pos1-2));
        int c = stoi(num2.substr(0,pos2)),d=stoi(num2.substr(pos2+1,m-pos2-2));
        int cnt1=a*c-b*d;
        int cnt2=a*d+c*b;
    return to_string(cnt1)+"+"+to_string(cnt2)+"i";
    }
};

I have even uglier code , What kind of human suffering is this

class Solution {
    
public:

    string complexNumberMultiply(string num1, string num2) {
    
        vector<string>ve;
        string xz="",xz1="",xz2="";
        int n=num1.size(),m=num2.size();
        int a=0,b=0,c=0,d=0;
        int flag1=0,flag2=0;
        int ans1=1,ans2=1;
        int cnt1=0,cnt2=0;
        char sss[100000];
        for(int i=0;i<n;++i){
    
            if(num1[i]>='0'&&num1[i]<='9'&&flag1==0){
    
                a=a*10+(num1[i]-'0');
            }
            else{
    
                if(i!=0){
    
                      if(num1[i]=='-'||num1[i]=='+'&&num1[i+1]=='-')
                ans1=-1;
                flag1=1;
                 if(num1[i]>='0'&&num1[i]<='9'){
    
                b=b*10+(num1[i]-'0');
                }
              
            }
            }
        }
        if(num1[0]=='-')
        a=-a;
        b*=ans1;
            for(int i=0;i<m;++i){
    
            if(num2[i]>='0'&&num2[i]<='9'&&flag2==0){
    
                c=c*10+(num2[i]-'0');
            }
            else{
    
                if(i!=0){
    
                    if(num2[i]=='-'||num2[i]=='+'&&num2[i+1]=='-')
                ans2=-1;
                flag2=1;
                 if(num2[i]>='0'&&num2[i]<='9'){
    
                d=d*10+(num2[i]-'0');
                }  
            }
            }
        }
        if(num2[0]=='-')
        c=-c;
        d*=ans2;
        cnt1=a*c-b*d;
        cnt2=a*d+c*b;
        cout<<a<< " "<<b<<" "<<c<<" "<<d<<endl;
        cout<<a*d<<" "<<c*b<<endl;
    cout<<cnt1<<" "<<cnt2<<endl;
        int sum1=0,sum2=0;
        string q1;
      
        if(cnt1<0){
    
            xz+="-";
            cnt1=-cnt1;
        }
          if(cnt1==0){
    
            xz+="0";
        }
        else{
    
            while(cnt1){
    
                int nnn=cnt1%10;
                   if(nnn==0)
                    q1="0";
                     if(nnn==1)
                    q1="1";
                     if(nnn==2)
                    q1="2";
                     if(nnn==3)
                    q1="3";
                     if(nnn==4)
                    q1="4";
                     if(nnn==5)
                    q1="5"; 
                    if(nnn==6)
                    q1="6";
                     if(nnn==7)
                    q1="7";
                     if(nnn==8)
                    q1="8";
                     if(nnn==9)
                    q1="9";
                 xz1+=q1;
                 cnt1/=10;
            }
            for(int j=xz1.size()-1;j>=0;--j)
            xz+=xz1[j];
        } 
         xz+="+";
        if(cnt2!=0){
     
            if(cnt2<0){
    
             xz+="-";
                cnt2=-cnt2;
            }
            string q;
            while(cnt2){
    
                int nnn1=cnt2%10;
                if(nnn1==0)
                    q="0";
                     if(nnn1==1)
                    q="1";
                     if(nnn1==2)
                    q="2";
                     if(nnn1==3)
                    q="3";
                     if(nnn1==4)
                    q="4";
                     if(nnn1==5)
                    q="5"; 
                    if(nnn1==6)
                    q="6";
                     if(nnn1==7)
                    q="7";
                     if(nnn1==8)
                    q="8";
                     if(nnn1==9)
                    q="9";
                 xz2+=q;
                 cnt2/=10;
            } 
             for(int j=xz2.size()-1;j>=0;--j)
            xz+=xz2[j];
        }else{
    
          xz+="0";
        }
        xz+="i";
       return xz;
    }
};

If you try , But it still has no effect , It shows that you are atoning for your previous laziness , As long as you stick to it, fight your head and blood , After this time, you will find , You climbed to an unimaginable height