Remove Duplicate Letters

public class Solution {
    public String removeDuplicateLetters(String s) {
        if (s == null || s.length() == 0) return "";
        
        int[] counts = new int[26];
        Stack<character> stack = new Stack<character>();
        for (char c : s.toCharArray()) {
            counts[c - 'a'] ++;
        }
        for (char c : s.toCharArray()) {
            counts[c - 'a'] --;
            if (stack.contains(c)) {
                continue;
            }
            while (!stack.isEmpty() && stack.peek() >= c && counts[stack.peek() - 'a'] > 0) {
                stack.pop();
            }
            stack.push(c);
        }
        StringBuilder sb = new StringBuilder();
        while (!stack.isEmpty()) {
            sb.append(stack.pop());
        }
        return sb.reverse().toString();
    }
}

public class Solution {
    //Given the string s, the greedy choice (i.e., the leftmost letter in the answer) 
    //is the smallest suffix s[i .. ] contains all the unique letters.
    //怎么挑第一个字母？先找到最小的包含所有unique letters的前缀，例如"cbacdcbc",最小的前缀是"cba",
    //第一个字母必须从这其中挑，因为后面的字符串不包含a了。在这个前缀里挑最小的字母，就是'a'，然后
    //移除a前面所有的字母，并去掉后面所有的a。
    public String removeDuplicateLetters(String s) {
        if (s == null || s.length() == 0) return "";
        
        int[] counts = new int[26];
        boolean[] candidates = new boolean[26];
        for (char c : s.toCharArray()) {
            counts[c - 'a'] ++;
        }
        for (char c : s.toCharArray()) {
            candidates[c - 'a'] = true;
            if (--counts[c - 'a'] == 0) {
                break;
            }
        }
        for (char c = 'a'; c <= 'z'; c ++) {
            if (candidates[c - 'a']) {
                for (int i = 0; i < s.length(); i++) {
                    if (s.charAt(i) == c) {
                        return c + removeDuplicateLetters(s.substring(i+1).replaceAll("" + c, ""));
                    }
                }
            }
        }
        return "";
    }
}