451. Sort Characters By Frequency

Posted 2020-10-08 apanda009

Given a string, sort it in decreasing order based on the frequency of characters.

Example 1:

Input:
"tree"

Output:
"eert"

Explanation:
‘e‘ appears twice while ‘r‘ and ‘t‘ both appear once.
So ‘e‘ must appear before both ‘r‘ and ‘t‘. Therefore "eetr" is also a valid answer.

Example 2:

Input:
"cccaaa"

Output:
"cccaaa"

Explanation:
Both ‘c‘ and ‘a‘ appear three times, so "aaaccc" is also a valid answer.
Note that "cacaca" is incorrect, as the same characters must be together.

Example 3:

Input:
"Aabb"

Output:
"bbAa"

Explanation:
"bbaA" is also a valid answer, but "Aabb" is incorrect.
Note that ‘A‘ and ‘a‘ are treated as two different characters.

The logic is very similar to NO.347 and here we just use a map a count and according to the frequency to put it into the right bucket. Then we go through the bucket to get the most frequently character and append that to the final stringbuilder.

public String frequencySort(String s) {
        
        HashMap<Character, Integer> map = new HashMap<>();
        for (char c : s.toCharArray()) {
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        ArrayList<Character>[] list = new ArrayList[s.length() + 1];
        for (char c : map.keySet()) {
            int frequency = map.get(c);
            
            if (list[frequency] == null) {
                list[frequency] = new ArrayList<>();
            }
            list[frequency].add(c);
        }
        StringBuilder sb = new StringBuilder();
        for (int i = s.length(); i >= 0; i--) {
            if (list[i] != null) {
                for (char c : list[i]) {
                    for (int j = 0 ; j < (int)map.get(c); j++) {
                        sb.append(c);
                    }
                    
                }
            }
        }
        return sb.toString();
    }

　　

public String frequencySort(String s) {
        Map<Character, Integer> map = new HashMap<>();
        for (char c : s.toCharArray()) {
            if (map.containsKey(c)) {
                map.put(c, map.get(c) + 1);
            } else {
                map.put(c, 1);
            }
        }
        PriorityQueue<Map.Entry<Character, Integer>> pq = new PriorityQueue<>(
            new Comparator<Map.Entry<Character, Integer>>() {
                @Override
                public int compare(Map.Entry<Character, Integer> a, Map.Entry<Character, Integer> b) {
                    return b.getValue() - a.getValue();
                }
            }
        );
        pq.addAll(map.entrySet());
        StringBuilder sb = new StringBuilder();
        while (!pq.isEmpty()) {
            Map.Entry e = pq.poll();
            for (int i = 0; i < (int)e.getValue(); i++) {
                sb.append(e.getKey());
            }
        }
        return sb.toString();
    }