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.

 

首先用hashmap扫一遍string，统计频率。按频率排序，将hashmap的entry存入max heap，再一个个poll出来，append到StringBuilder

时间：O(NlogN)，空间：O(N)

class Solution {    public String frequencySort(String s) {        HashMap
     
       map = new HashMap<>();        PriorityQueue
      
       
        > maxHeap = new PriorityQueue<>((a, b) -> (b.getValue() - a.getValue()));                for(char c : s.toCharArray()) {            map.put(c, map.getOrDefault(c, 0) + 1);        }                for(Map.Entry
        
          entry : map.entrySet()) {            maxHeap.offer(entry);        }                StringBuilder sb = new StringBuilder();        while(!maxHeap.isEmpty()) {            Map.Entry
         
           tmp = maxHeap.poll();            for(int j = 0; j < tmp.getValue(); j++)                sb.append(tmp.getKey());        }        return sb.toString();    }}
         
        
       
      
     

class Solution {    public String frequencySort(String s) {        Map
     
       map = new HashMap<>();        for(int i = 0; i < s.length(); i++) {            map.put(s.charAt(i), map.getOrDefault(s.charAt(i), 0) + 1);        }                PriorityQueue
      
       
        > maxHeap = new PriorityQueue<>((a, b) -> b.getValue() - a.getValue());        maxHeap.addAll(map.entrySet());                StringBuilder sb = new StringBuilder();        while(!maxHeap.isEmpty()) {            Map.Entry
        
          entry = maxHeap.poll();            for(int i = 0; i < entry.getValue(); i++)                sb.append(entry.getKey());        }        return sb.toString();    }}