字符串，求教,该如何处理_J2SE

字符串，求教
.解析一个字符串，对字符串中重复出现的字符，只在第一次出现时保留
如: abdabbefgf 结果为 abdefg

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Java code

import java.util.ArrayList;import java.util.List;import java.util.Scanner;public class Test2 {    public static void main(String[] args){                Scanner sc = new Scanner(System.in);        String str = sc.next();        List<Character> list = new ArrayList<Character>();        for(int i=0; i<str.length(); i++){            if(!list.contains(str.charAt(i))){                list.add(str.charAt(i));            }        }        System.out.println(list);    }}
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Java codeimport java.util.LinkedHashSet;public class Test{    public static void main(String args[]){        LinkedHashSet<Character> set = new LinkedHashSet<Character>();        String s = "abdabbefgf";        for(Character c:s.toCharArray()){            if(set.add(c)){            System.out.print(c);                }        }        System.out.println(set);    }}
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for example
Java codeString s = "abdabbefgf";StringBuilder buf = new StringBuilder();char[] c = s.toCharArray();for (int i=0; i<c.length; i++) {    if (s.indexOf(c[i]) == i) buf.append(c[i]);}System.out.println(buf);