原题链接在这里：

题目：

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.

题解：

类似. 利用Bucket Sort 按照频率把s的char放进bucket里, 再按照频率从大到小，重复频率次append到res中.

Time Complexity: O(n), n = s.length(). Space: O(n).

AC Java:

1 public class Solution { 2     public String frequencySort(String s) { 3         if(s == null || s.length() == 0){ 4             return s; 5         } 6          7         HashMap
     
       freqMap = new HashMap
      
       (); 8         for(char c : s.toCharArray()){ 9             freqMap.put(c, freqMap.getOrDefault(c, 0)+1);10         }11         12         StringBuilder [] bucket = new StringBuilder[s.length()+1];13         for(char c : freqMap.keySet()){14             int freq = freqMap.get(c);15             if(bucket[freq] == null){16                 bucket[freq] = new StringBuilder();17             }18             for(int i = 0; i
       
        =0; i--){25             if(bucket[i] != null){26                 res.append(bucket[i]);27             }28         }29         return res.toString();30     }31 }