Character set (de duplication)

Enter a string , Find the character set contained in the string , Output in alphabetical order .

Data range ： The length of the input string meets the requirements 1≤n≤100 1 \le n \le 100 \ 1≤n≤100 , And contains only uppercase and lowercase letters , Case sensitive .

There are many groups of input

Input ：

 Enter a string for each group of data , The maximum length of the string is 100, And contains only letters , It can't be an empty string , Case sensitive .

Output ：

 One line for each group of data , In the original character order of the string , Output character set , That is, the repeated and subsequent letters are not output .

First of all, the input of this question is a string composed of upper and lower case letters with only letters . Output non repeating elements in order .

Topic idea ：

First , Each of our letters has a relative int An array of types . We just need to traverse this string , Find the value of each character , At the same time , If the value of the array corresponding to this character is 0 Then it means there is no repetition , Then add this character to another reference variable , Then we'll talk about the value of the array corresponding to this character ++, So if there's a duplicate letter after it , Every time I find this array , The discovery is not 0, Then it won't be added to string In the reference . So we can solve the problem .

import java.util.Scanner;

public class Main {
    public static StringBuilder func(String str){
        int[] ch = new int[257];
        StringBuilder set = new StringBuilder("");
        int tem;
        for(int i = 0;i<str.length();i++){
            tem = str.charAt(i);
            if(ch[tem]==0){
                set.append(str.charAt(i));
                ch[tem]++;
            }

        }
        return set;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String str = scanner.nextLine();
        StringBuilder set = func(str);
        System.out.println(set.toString());
    }
}

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