to find repetative character in string

How can I find distin开发者_StackOverflow中文版ct repetitive character in string using Java.

For the string 4567895443577

Here, the first distinct repetitive character is 5

Ip:n:1 output:4
   n=2     op=5
   n=3     op=7
   n=4     op=doest exist

create HashSet and HashMap: set,map and int count=0, iterate over the string, and add each character and its index. at the end - each character's value will be the LAST index.
iterate over the String again, and check if the index is as appears in the map. if it does (or the character appears in the set) - ignore it.
if a character is not in the set, and index as is and as in map don't match - increase count (until it reaches n).

complexity: O(n)

public static Character findN(String str,int n) { 
    HashMap<Character, Integer> map = new HashMap<Character, Integer>();
    int len = str.length();
    for (int i=0;i<len;i++) { 
        map.put(str.charAt(i),i);
    }
    int count=0;
    HashSet<Character> set = new HashSet<Character>();
    for (int i=0;i<len;i++) {
        if (set.contains(str.charAt(i))) continue;
        if (map.get(str.charAt(i)) != i) {
            count++;
            if (count == n) return str.charAt(i);
            set.add(str.charAt(i));
        }
    }
    return null; //it does not exist

}

This should work:

public static char findChar(String s, int length) {
int[] counts = new int[10];

// iterate over the letters and increment the count
int stringLength = s.length();
for(int i = 0; i < stringLength; i++ ) {
    char c = s.charAt(i);
    int value = Character.getNumericValue(c);
    counts[value]++;
}

int counter = 0; // how many chars repeated so far
for(int i = 0; i < stringLength; i++ ) {
    char c = s.charAt(i);
    int value = Character.getNumericValue(c);
    if(counts[value] >= 2) {

    counts[value] = -1; // do not count this twice
    counter++;

    if(counter == length) {
        return c;
    }
    }
}
return '\u0000'; // null char
}

This can be done by the following code.

I have used HashMap keys as an input characters and value as a counter.

String str = "4567895443577";
char[] chars = str.toCharArray();
HashMap<Character, Integer> charMap = new HashMap<Character, Integer>();
for( char c : chars )
{
    if( charMap.containsKey( c ) ){
        charMap.put(c, charMap.get(c) + 1 );
    }else{
        charMap.put(c, 1);
    }
}
for( Entry<Character, Integer> entry : charMap.entrySet() )
{
    System.out.println( "Character '"+entry.getKey()+"' is repeated for '"+entry.getValue()+"' times." );
}

You should create a HashSet which implements Set Interface.

A collection that contains no duplicate elements. More formally, sets contain no pair of elements e1 and e2 such that e1.equals(e2), and at most one null element. As implied by its name, this interface models the mathematical set abstraction.

 /* 
 * Ex-OR basic : 0^0 = 0, 0^1 = 1, 1^0 = 1, 1^1 = 0  
 * 
  Ex-ORing bits of all characters in String nums = "4567895443577"
  i   Operation                Bitwise operation    Result(bin)   Result(Dec)  
  0       4 ^ 5 ...arr[0]^arr[1]       100 ^ 101            001             1                 
  //NOTE : first occurence as result = 1 should be skipped
  ----------------------------------------------------------------------------      
       Result(i-1)        arr[i]
  for:  
  1       1        ^        5          001 ^ 101            100             4
  2       4        ^        6          100 ^ 110            010             2
  3       2        ^        7          010 ^ 111            101             5
  4       5        ^        8          0101 ^ 1000         1101            13
  5      13        ^        9          1101 ^ 1001         0100             4
  6       5        ^        4          0101 ^ 0100         0001             1                  
 // break "for" found repeated element. return 5           
 * */
public class RepeatedNumber {
 public static void main(String args[]) {
    String nums = "4567895443577";
    char  repeated = (char) findRepeated(nums.toCharArray()) ;
    System.out.println("result ="+repeated);
 }
 public static int findRepeated(char arr[]) {

    int result = arr[0]^arr[1];
    int repeated = arr[0];
    //find out number repeated more than once in array
    if(result != 0) {
        for(int i = 1; i < arr.length; i++) {

            result = result ^ arr[i];

            if(result == 1 || arr[i] == arr[i-1]) {

                repeated = arr[i];
                break;
            }
        }
    }
    return repeated;
 }
}