Q:

Eliminate duplicates from strings

I have written a piece of code for removing duplicated values from a string. Any suggestions on improvements?
public static String removeDuplicate(String s)
{
    char [] temp = s.toCharArray();
    int length =temp.length; 
    for (int i=0;i<length;i++)
    {
        for (int j = i+1; j<length;j++)
        {
            if(temp[i]==temp[j])
            {
                int test =j;
                for(int k=j+1; k<length ; k++)
                {
                    temp[test] = temp[k];
                    test++;
                }
                length--;
                j--;
            }
        }
    }
    return String.copyValueOf(temp).substring(0,length);
}

A:

Yusshi's code is perfectly fine. If you know that all the characters are in a specific range (e.g., ASCII code 0 to 255), you could also use either of the following codes which is faster. They uses an array rather than a hash set. The time complexity is O(n) and the space complexity is O(1). You cannot go lower in either case.
This one sorts the characters in the output string:
public static String removeDuplicates(String str) {
    int charsCount[] = new int[256];

    for (int i = 0; i < str.length(); i++) {
        char ch = str.charAt(i);
        charsCount[ch]++;
    }

    StringBuilder sb = new StringBuilder(charsCount.length);
    for (int i = 0; i < charsCount.length; i++) {
        if (charsCount[i] > 0) {
            sb.append((char)i);
        }
    }

    return sb.toString();
}

This one preserves the order of the characters:
public static String removeDuplicates(String str) {
    boolean seen[] = new boolean[256];
    StringBuilder sb = new StringBuilder(seen.length);

    for (int i = 0; i < str.length(); i++) {
        char ch = str.charAt(i);
        if (!seen[ch]) {
            seen[ch] = true;
            sb.append(ch);
        }
    }

    return sb.toString();
}

A:

This can be simplified a great deal with an auxiliary data structure. Basically, we want to scan the string, and only add characters that we have not yet seen. Hence, if we use a data structure with fast insert/lookup, we can quickly tell if we've already seen that character without having to rescan over the string. In this case, this suggests a Set:
import java.util.Set;
import java.util.HashSet;

public static String removeDuplicate(String s)
{
    StringBuilder sb = new StringBuilder();
    Set<Character> seen = new HashSet<Character>();

    for(int i = 0; i < s.length(); ++i) {
        char c = s.charAt(i);
        if(!seen.contains(c)) {
            seen.add(c);
            sb.append(c);
        }
    }
    return sb.toString();
}

This trades time for space, so in this case, it should make the algorithm \$O(N)\$ time (as opposed to \$O(N^3)\$) at the cost of using \$O(N)\$ extra space.

A:

Style
Java Code-style conventions put:

the { brace at the end of the line that introduces the compound statement block, not on the next line.
spaces between variables and operators: for (int i=0;i<length;i++)

Your code should look like:

public static String removeDuplicate(String s) {
    char[] temp = s.toCharArray();
    int length = temp.length;
    for (int i = 0; i < length; i++) {
        for (int j = i + 1; j < length; j++) {
            if (temp[i] == temp[j]) {
                int test = j;
                for (int k = j + 1; k < length; k++) {
                    temp[test] = temp[k];
                    test++;
                }
                length--;
                j--;
            }
        }
    }
    return String.copyValueOf(temp).substring(0, length);
}

Algorithm
You have three nested loops. This is inefficient.
A slight nit-pick is that you also use an inefficient way to get the resulting String using the copyValueOf(...) and then the substring(...).
I believe if you take an 'additive' approach to the process you will have a faster system. By extracting the onle loop in to a function it is also more readable:
private static final boolean foundIn(char[] temp, int size, char c) {
    for (int i = 0; i < size; i++) {
        if (temp[i] == c) {
            return true;
        }
    }
    return false;
}

public static String removeDuplicateY(String s) {
    char[] temp = s.toCharArray();
    int size = 0; //how many unique chars found so far
    for (int i = 0; i < temp.length; i++) {
        if (!foundIn(temp, size, temp[i])) {
            // move the first-time char to the end of the return value
            temp[size] = temp[i];
            size++;
        }
    }
    return new String(temp, 0, size);
}

