1. Find the number that only appears once in the array , Other elements appear twice .

Method 1


package example;
/*
*@author Wang Zhihua
*@E-mail: [email protected]
*@date Creation time :2018 year 1 month 26 day afternoon 2:42:37
**/
public class FindOne {
    public static void main(String[] args) {
        int[] a={2,2,4,5,9,9,4};
        int t=0;
        for (int i = 0; i < a.length; i++) {
            t^=a[i];// Use exclusive or (^) The ingenious use of budget : Get a sum of numbers 0 Exclusive or self
            // A number is different from itself 0 Principle of , And the exclusive or operation can be exchanged , So the final result is the number of times
        }
        System.out.println(t);

    }


}

Method 2

     utilize ASCII Code creates a subscript to 256 Array of , The initial values are all assigned as 0, Then traverse the array Characters appear every time array value is added 1 After traversal
Number organization is n That's what happened n second , This method can calculate the number of characters

Or as follows


Each character corresponds to one anyway ASCII code , Count the subscript of each character .
need 2 Special constants 2 Middle case : 1, This character doesn't appear ( In the standard process 0) 2, This character appears more than once ( In the standard process -1)


 private static char firstAppearsOnlyonce(String str) {
            int[] hash = new int[256];                 
// Record the number of each character , share 256 individual ASCII code
            for(int i=0; i<256; i++){
                hash[i] = 0;
            }
            for(int i=0; i<str.length(); i++){
                hash[str.charAt(i)] ++;                 // Establish a reflection relation between a character and its number !
            }
            for(int i=0; i<str.length(); i++){           // Traverse the string again , Find the first occurrence of a character
                if(hash[str.charAt(i)] == 1){
                    return str.charAt(i);
                }
            }


2 Given an English string , Please write a piece of code to find the English character that appears three times in this string .
Enter a description :
Input data a string , Include letters , Number, etc .

package example;
/*
*@author Wang Zhihua
*@E-mail: [email protected]
*@date Creation time :2018 year 1 month 26 day afternoon 2:42:37
**/
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
/*
 * Given an English string , Please write a piece of code to find the English character that appears three times in this string .
Enter a description :
Input data a string , Include letters , Number, etc .
 *
 */

public class FindChar {
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        Map<Character,Integer> map = new HashMap<Character,Integer>();
        System.out.println(" Please enter the string you want to test ");
        String s=in.next();
        char[] c=s.toCharArray();
        for (int i = 0; i < c.length; i++) {
                if(!(c[i]>='a'&&c[i]<='z') && !(c[i]>='A'&&c[i]<='z'))
                                continue;
                   if(!map.containsKey(c[i])){
                          map.put(c[i], 1);// If ,map Does not contain this element , For each element value Set to 1
                            }
                            else{
                                int val = map.get(c[i])+1;// If you already have child elements
Of this element valve+1
                                if(val == 3){
                                    System.out.println(c[i]);
                                    break;
                                }
                                map.put(c[i],val);                         
         
        }        
        }
    }
    }


3, Print Yanghui triangle

package yanghui;

/*
*@author Wang Zhihua
*@E-mail: [email protected]
*@date Creation time :2018 year 1 month 26 day afternoon 2:42:37
**/


import java.util.Scanner;
public class PrintYanghui {
    public static void main(String[] args) {
        Scanner in =new Scanner(System.in);
        System.out.println(" Please input the number of lines you want to print Yanghui triangle ");
        int num=in.nextInt();
        int[][] a=new int[num][num];
        for (int i = 0; i <a.length; i++) {
            for (int j = 0; j <=i; j++) {
                 if(j==0||j==i){// Find rule assignment , Remember not to use the number of lines i=0; because i=0 Only once ,j=0 Multiple times can be obtained through internal circulation
                     a[i][j]=1;
                 }else{
                a[i][j]=a[i-1][j]+a[i-1][j-1];
                 }    
            }
        }
        for (int i = 0; i < a.length; i++) {
            for (int j = 0; j < 8-i; j++) {
                System.out.print(" ");
            }
            for (int j = 0; j <=i; j++) {
                System.out.print(a[i][j]+" ");
            }
            System.out.println("\n");    
        }
        }
}    

4. produce N Random number , And the number of digits is 1,2,3,4,5,6,7,8,9 Number of
Method 1 : Record the number of different digits with nine variables

package example1;
/*
*@author Wang Zhihua
*@E-mail: [email protected]
*@date Creation time :2018 year 1 month 26 day afternoon 2:42:37
**/
import java.util.Random;
import java.util.Scanner;

public class random {

    public static void main(String[] args) {
        Random r=new Random();
        Scanner in=new Scanner(System.in);
        System.out.println(" Please enter the number of random numbers N");
        int N=in.nextInt();
        int[] a=new int[N];
        int sum=0;    
        int sum1=0;    
        int sum2=0;    
        int sum3=0;    
        int sum4=0;    
        int sum5=0;    
        int sum6=0;
        int sum7=0;
        int sum8=0;
        int sum9=0;
        int m=0;
            
for (int i = 0; i < a.length; i++) {
    a[i]=r.nextInt(100);
}
System.out.println(" The resulting random number is :");
for (int i = 0; i < a.length; i++) {
    System.out.print(a[i]+" ");
        }
        for (int i = 0; i < a.length; i++) {
            m=a[i]-((a[i]/10)*10);
            if (m==0) {
                sum++;
            }else if (m==1){
                sum1++;
            }
        else if (m==2){
            sum2++;
        }
    else if (m==3){
        sum3++;
    }
else if (m==4){
    sum4++;

}else if (m==5){
    sum5++;

}else if (m==6){
    sum6++;

}else if (m==7){
    sum7++;

}else if (m==8){
    sum8++;
}
else if (m==9){
    sum9++;
}
}
System.out.println(" The number of digits is 0 The total number of is :"+sum);
System.out.println(" The number of digits is 1 The total number of is :"+sum1);
System.out.println(" The number of digits is 2 The total number of is :"+sum2);
System.out.println(" The number of digits is 3 The total number of is :"+sum3);
System.out.println(" The number of digits is 4 The total number of is :"+sum4);
System.out.println(" The number of digits is 5 The total number of is :"+sum5);
System.out.println(" The number of digits is 6 The total number of is :"+sum6);
System.out.println(" The number of digits is 7 The total number of is :"+sum7);
System.out.println(" The number of digits is 8 The total number of is :"+sum8);
System.out.println(" The number of digits is 9 The total number of is :"+sum9);

    }

}

Method 2 : Using arrays x[] Record the number of different digits , Obviously, the amount of code is much smaller , Therefore, arrays are also marked with the same records as variables ;

/*
*@author Wang Zhihua
*@E-mail: [email protected]
*@date Creation time :2018 year 1 month 26 day afternoon 2:42:37
**/


public class random {
    public static void main(String[] args) {
        Random r=new Random();
        Scanner in=new Scanner(System.in);
        System.out.println(" Please enter the number of random numbers N");
        int N=in.nextInt();
        int[] a=new int[N];
        int[] x=new int[9];
        int m=0;
        for (int i = 0; i < x.length; i++) {
            x[i]=0;// Initialize array as 0
        }
            
for (int i = 0; i < a.length; i++) {
    a[i]=r.nextInt(100);
}
System.out.println(" The resulting random number is :");
for (int i = 0; i < a.length; i++) {
    System.out.print(a[i]+" ");
        }
        for (int i = 0; i < a.length; i++) {
            m=a[i]-((a[i]/10)*10);
            if (m==0) {
                x[0]++;
            }else if (m==1){
                x[1]++;
            }
        else if (m==2){
            x[2]++;
        }
    else if (m==3){
        x[3]++;
    }
else if (m==4){
    x[4]++;

}else if (m==5){
    x[5]++;

}else if (m==6){
    x[6]++;

}else if (m==7){
    x[7]++;

}else if (m==8){
    x[8]++;
}
else if (m==9){
    x[9]++;
}
}
        for (int i = 0; i < x.length; i++) {
            System.out.println(" The number of digits is 0 The total number of is :"+x[i]);
        }}
    }
}

5. Realize the function of sending red packets via wechat , Enter amount m, Number of red packets n, produce n Random number , cut n The sum of the red packets is m

package example3;
import java.math.BigDecimal;
import java.util.Random;
import java.util.Scanner;
public class RedPacket {
    public static void main(String[] args) {
    Scanner in=new Scanner(System.in);
    String choice;
    do{
    System.out.println(" Please input the number of red packets you want to send ");
    int n=in.nextInt();
    System.out.println(" Please enter the total amount of red packets you want to send ");
    double sum=in.nextDouble();
    double []  money=new double[n];
    for (int i = 0; i < money.length-1; i++) {
                money[i]=random(sum);
                sum-=money[i];
            }
              money[money.length-1]=sum;
        for (int i = 0; i < money.length; i++) {
            double d=money[i];
            BigDecimal mData = new BigDecimal(d).setScale(2,
BigDecimal.ROUND_HALF_UP);
            // Accurate to Two decimal places
            System.out.println(mData);           
            }
         System.out.println(" Do you want to continue with the program :Y: continue     N end ");
         choice=in.next();
                }while(choice.equals("Y"));
}
    public static double random(double money){
            double m;
            Random r=new Random();
            m= r.nextDouble();
            if (m<=0) {
                m= r.nextDouble();
            }
            
            return m*money;
}

}

6. realization vip Login function Every day 3 Input error opportunities

package example2;

/*
*@author Wang Zhihua
*@E-mail: [email protected]
*@date Creation time :2018 year 1 month 26 day afternoon 2:42:37
**/

import java.util.Scanner;
public class isVip {
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner in = new Scanner(System.in);
        System.out.println(" Please enter your account number ");
        String s = in.next();
        int a = 60;
        int m = 3;
        int x = 0;

        do {
            if (!s.equals("888888")) {
                --m;
                if(m>0){
                System.out.println(" You still have " + m + " Opportunities ");
                System.out.println(" Please re-enter your account ");
                s = in.next();}
                if (m == 1) {
                    x = 1;
                }
            } else {
                break;
            }
        } while (m > 0);
        if (x == 1) {
            System.out.println(" bye ");
        } else {
            System.out.print(" Welcome vip");
        }

    }

}
7. seek s=a+aa+aaa+aaaa+aa...a Value of , among a It's a number . For example, input 2 output 2+22+222+2222+22222+222222=246912
 package example2;

/*
*@author Wang Zhihua
*@E-mail: [email protected]
*@date Creation time :2018 year 1 month 26 day afternoon 2:42:37
**/import java.util.Scanner
public class deno {
    public static void main(String[] args) {
        int sum=0;
        Integer m=0;
        String str1="";
        Scanner in=new Scanner(System.in);
        System.out.println(" Please enter an integer ");
        int a=in.nextInt();
        for (int i = 1; i <=a; i++) {
             m+=Integer.valueOf(solution(i,a));
            sum+=m;
        }
        for (int i = 1; i <=8; i++) {
            str1+=solution(i,a)+"+";
        }
        System.out.println(str1+"="+m);
        
    }
 
  public static String  solution(int a,int b){
      String str="";
    for (int i = 0; i < a; i++) {
        str+=b;// Evaluate or string with string and any type
    }
    return str;
  }
}
8. Given an array , Find out the elements whose total number of the same elements is greater than half the size of the array , And output

package example4;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
package example4;
/*
*@author Wang Zhihua
*@E-mail: [email protected]
*@date Creation time :2018 year 1 month 26 day afternoon 7:25:26
* /
public class FindMax {
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        Map<Integer,Integer> map=new HashMap();
        System.out.println(" Please enter the size of the array you want to test ");
        int n=in.nextInt();
        int[] a=new int[n];
        int i;
        System.out.println(" Please enter the elements of the array you want to test ");
        for ( i = 0; i < a.length; i++) {
            int m=in.nextInt();
            a[i]=m;
            map.put(a[i], 1);
        boolean b=map.containsKey(a[i]);
            if (b) {
                int p=map.get(a[i])+1;// If the same element appears ,key Corresponding valve add one-tenth
                map.put(a[i], p);
            }
        }
            for (int j = 0; j < map.size(); j++) {
                if (map.get(a[j])>=a.length/2) {
                    System.out.println(a[j]);
                }else {
                    System.out.println(" non-existent ");
                }
            }
    }
}
9. Write a two color ball prize system ( The classic application of tag array )
/*
*@author Wang Zhihua
*@E-mail: [email protected]q.com
*@date Creation time :2018 year 1 month 27 day afternoon 7:00:43
*            Dichroic sphere
         Dichroic demand : Choose six red balls , Choose a basketball player
             Red ball 6 individual :1--33
             Blue ball 1 individual :1--16            
        1. Build an array of red balls 33 Elements , Each element is 1.2.3.4.5.....33                        
        2. Random sampling 6 Balls in an array
        3. Blue ball 1 individual
**/
package array.douublecolorball;
import java.util.Random;

public class DouubleColorBall {
    public static void main(String[] args) {
        int[] reaball=new int[33];
        String result="";
        int[] luckRadball=new int[6];
        int luckBlueBall=0;
        boolean[] b=new boolean[33];
        Random r=new Random();
        luckBlueBall=r.nextInt(16)+1;
        for (int i = 0; i < reaball.length; i++) {
            reaball[i]=i+1;
        }
        for (int j = 0; j < luckRadball.length; j++) {
            if (!b[j]) {// If the random number is not in the array , here b[i] Is the default false
                luckRadball[j]=r.nextInt(33)+1;
                b[j]=true;// At this time, the random number appears in the array , hold b[i] Mark as true
            }else {// In other cases, random numbers are generated uniformly, and they are not repeated
                do {
                    luckRadball[j]=r.nextInt(33)+1;
                } while (b[j]=true);
            }
        }
        for (int i = 0; i < luckRadball.length; i++) {
            result+=luckRadball[i]+" ";
        }
        System.out.println(" The result of this double color ball is :"+" Red ball : "+result+" Blue ball : "+luckBlueBall)
    }
}

9 seek 100 reach 1000 Number of daffodils within
/*
*@author Wang Zhihua
*@E-mail: [email protected]
*@date Creation time :2018 year 1 month 27 day afternoon 7:50:27
**/

package array.flowerNumber;

public class FlowerNumber {

    public static void main(String[] args) {  
        System.out.println("100-1000 The number of daffodils in :");  
        for(int i=100;i<1000;i++){  
            int ge  = i%10;  
            int shi = i/10%10;  
            int bai = i/10/10%10;  
              
            // Judgment requirements for number of Narcissus  
            if(i == (ge*ge*ge+shi*shi*shi+bai*bai*bai)){  
                System.out.println(i);  
            }  
        }
 
    }
}

10. Given an integer , Find its binary number , And input .

/*
*@author Wang Zhihua
*@E-mail: [email protected]
*@date Creation time :2018 year 1 month 28 day afternoon 3:30:47
**/

package tentotwo;
import java.util.Scanner;
public class TenToTwo {

    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        System.out.println(" Please enter an integer ");
        int num=in.nextInt();
        int num1=Math.abs(num);// Use its absolute value first, both positive and negative
        StringBuffer str1=new StringBuffer();    
        while(num1/2!=0||num1%2!=0){        
            str1.append(num1%2+"");
            num1=num1/2;        
        }
        if (num<0) {
            str1.append(1+"");// If it's a negative number , Add a sign bit before 1, Plus or minus
        }
        StringBuffer str=str1.reverse();// Because it's from the back to the front , So we need to invert the string
        System.out.println(str);
    }
}

11. Enter a string , Find the number string with the longest consecutive number in the string , as input as1254dad14  output 1254


/* Algorithmic idea : use max Represents the maximum number length passed ,count Represents a digital counter , Reset to when letter 0
*end Represents the end of a number , Every time the number is met , Yes max Make a judgment , When max Less than count Time , to update max and end
*/


/*
*@author Wang Zhihua
*@E-mail: [email protected]
*@date Creation time :2018 year 1 month 28 day afternoon 12:57:17
**/
package maxstring;
import java.util.Scanner;
public class MaxStr {
    public static void main(String[] args) {
        // TODO Auto-generated method stub
Scanner scanner = new Scanner(System.in);
          while(scanner.hasNext()){
              String str = scanner.nextLine();
              int max = 0;
              int count=0,end=0;
              for(int i=0;i<str.length();i++){
                  if(str.charAt(i)>='0' && str.charAt(i)<='9'){
                      count++;
                      if(max<count){
                          max= count;
                          end = i;
                      }
                  }else{
                      count = 0;
                  }
              }
              System.out.println(str.substring(end-max+1,end+1));
          }
}