以递归方式从字符串中删除重复的字符

Question

我需要帮助来确定如何从字符串中删除重复字符.它必须以递归方式完成,这才是真正的问题.

public class FEQ2 {
    /**
     * @param args
     */
    public static void removeDups(String s, int firstChar, int secondChar) {    
        if (s.length() == 1) {  
            System.out.println(s);
        }           
        char a = s.charAt(firstChar);
        if (a == s.charAt(secondChar)) {
            s = a + s.substring(secondChar + 1);
        }
        System.out.println(s);
        removeDups(s, firstChar + 1, secondChar + 1);
        //return s;
    }

    public static void main(String[] args) {
        //System.out.println(removeDups("AAAABBARRRCC", 1));
        removeDups("AAAABBARRRCC", 0 , 1);
    }
}

Answer 1

你可以这样做:

public static String removeDups(String s)
{
    if ( s.length() <= 1 ) return s;
    if( s.substring(1,2).equals(s.substring(0,1)) ) return removeDups(s.substring(1));
    else return s.substring(0,1) + removeDups(s.substring(1));
}


INPUT: "AAAABBARRRCC"
OUTPUT: "ABARC"

===============

编辑:另一种方式

public static String removeDups(String s)
{
    if ( s.length() <= 1 ) return s;
    if( s.substring(1).contains(s.substring(0,1)) ) return removeDups(s.substring(1));
    else return s.substring(0,1) + removeDups(s.substring(1));
}


INPUT: "AAAABBARRRCC"
OUTPUT: "BARC"

==============

编辑:第三种方式

public static String removeDups(String s)
{
    if ( s.length() <= 1 ) return s;
    if( s.substring(0,s.length()-1).contains(s.substring(s.length()-1,s.length())) ) return removeDups(s.substring(0,s.length()-1));
    else return removeDups(s.substring(0,s.length()-1)) + s.substring(s.length()-1,s.length());
}


INPUT: "AAAABBARRRCC"
OUTPUT: "ABRC"