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. 用Bucket Sort解决问题。思路如下：1、创建hashmap存相应的字符，
2、创建bucket在相应的框里存字符，
3、后向前遍历添加字符串。
代码如下：
public class Solution { public String frequencySort(String s) { HashMap<Character, StringBuilder> hm = new HashMap<Character, StringBuilder>(); char[] chs = s.toCharArray(); int len = chs.length; StringBuilder[] strs = new StringBuilder[len + 1]; StringBuilder res = new StringBuilder(); for (char ch: chs) { if (!hm.containsKey(ch)) { StringBuilder temp = new StringBuilder(); hm.put(ch, temp); } hm.get(ch).append(ch); } for (char ch: hm.keySet()) { int lens = hm.get(ch).length(); if (strs[lens] == null) { strs[lens] = new StringBuilder(); } strs[lens].append(hm.get(ch)); } for (int i = len; i >= 0; i--) { if (strs[i] != null) { res.append(strs[i]); } } return res.toString(); } }