Java实现字符串全排列无重复解决方案

//字符串全排列问题 (解决了重复问题)
import java.util.*;   
public class ListALL { 
   static int count=0; 
   public static void main(String[] arg) { 
       Scanner r=new Scanner(System.in); 
       String s=r.nextLine(); 
       Pailie(s, " "); 
       System.out.println("共有 "+count+"种情况"); 
     } 
   static void Pailie(String s,String p)   { 
       if(s.length()<1)   { 
       System.out.println(p);//字符串长度小于1，换行 
       count++; 
       } 
       else{ 
          int index[]=new int[s.length()]; 
          for(int i=0;i<s.length();i++)
             //该循环将所有字符的第一次出现的位置记录在数组index中 
          index[i]=s.indexOf(s.charAt(i)); 
          for(int i=0;i<s.length();i++){ 
             if(i==index[i])
               //只有当循环数与第一次记录数相等时才递归，保证相同字符中的第一个调用 
             Pailie(s.substring(1),p+s.substring(0,1));//递归，打印其它字符 
             s=s.substring(1)+s.substring(0,1);//循环移位 
             } 
          } 
      } 
} 

/*import java.util.LinkedList;
public class ListALL{
private String input;
public ListALL(String input1) {
this.input = input1;
}
@SuppressWarnings("unchecked")
public LinkedList<LinkedList<Character>> insert(
LinkedList<LinkedList<Character>> list, char inserChar) {
LinkedList<LinkedList<Character>> reValue = new LinkedList<LinkedList<Character>>();
LinkedList<Character> temp1;
LinkedList<Character> temp2;
for (int i = 0; i < list.size(); i++) {
temp1 = list.get(i);
for (int j = 0; j <= temp1.size(); j++) {
temp2 = (LinkedList<Character>) temp1.clone();
// temp2 = copy(temp1);
temp2.add(j, inserChar);
reValue.add(temp2);
}
}
return reValue;
}

public LinkedList<LinkedList<Character>> paixu() {
char[] inputChar = this.input.toCharArray();
LinkedList<LinkedList<Character>> ini = new LinkedList<LinkedList<Character>>();
LinkedList<LinkedList<Character>> reValue = new LinkedList<LinkedList<Character>>();
LinkedList<Character> temp = new LinkedList<Character>();
temp.add(inputChar[0]);