[leetcode-316]Remove Duplicate Letters

该题思路有两种，一种是将string当做stack处理，另一种是贪婪算法

class Solution {
public:
    string removeDuplicateLetters(string s) {
        bool inRes[26] = {false};
        int counts[26] = {0};
        
        for (auto ch : s) counts[ch - 'a']++;
        
        string res = "";
        
        for (auto ch : s) {
            counts[ch - 'a']--;
            
            if (res.empty()) {
                res.push_back(ch);
                inRes[ch - 'a'] = true;
                continue;
            }
            
            if (inRes[ch - 'a']) {
                continue;
            }
            
            while (ch < res.back() && !res.empty() && counts[res.back() - 'a'] > 0) {
                inRes[res.back() - 'a'] = false;
                res.pop_back();
            }
            
            res.push_back(ch);
            
            inRes[ch - 'a'] = true;
        }
        
        return res;
    }
};

Given the string s, the greedy choice (i.e., the leftmost letter in the answer) is the smallest s[i], s.t. the suffix s[i .. ] contains all the unique letters. (Note that, when there are more than one smallest s[i]'s, we choose the leftmost one. Why? Simply consider the example: "abcacb".)

After determining the greedy choice s[i], we get a new string s' from s by 

1. removing all letters to the left of s[i],
2. removing all s[i]'s from s.

We then recursively solve the problem w.r.t. s'. 

The runtime is O(26 * n) = O(n).

public class Solution {
    public String removeDuplicateLetters(String s) {
        int[] cnt = new int[26];
        int pos = 0; // the position for the smallest s[i]
        for (int i = 0; i < s.length(); i++) cnt[s.charAt(i) - 'a']++;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) < s.charAt(pos)) pos = i;
            if (--cnt[s.charAt(i) - 'a'] == 0) break;
        }
        return s.length() == 0 ? "" : s.charAt(pos) + removeDuplicateLetters(s.substring(pos + 1).replaceAll("" + s.charAt(pos), ""));
    }
}