示例 1:
输入:"tree"
输出:"eert"
解释:'e'出现两次，'r'和't'都只出现一次。因此'e'必须出现在'r'和't'之前。此外，"eetr"也是一个有效的答案。
来源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/sort-characters-by-frequency著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

    public String frequencySort(String s) {        Map<Character, Integer> map = new HashMap<>();        for (char ch : s.toCharArray()) {            map.put(ch, map.getOrDefault(ch, 0) + 1);        }
        List<Character> list = new ArrayList<>(map.keySet());        Collections.sort(list, (a, b) -> map.get(b) - map.get(a));        StringBuffer ans = new StringBuffer();        for (char item : list) {            int frequency = map.get(item);            while (frequency > 0) {                ans.append(item);                frequency--;            }        }
        return ans.toString();    }

    public String frequencySort(String s) {        Map<Character, Integer> map = new HashMap<>();        int maxFreq = 0;        for (char ch : s.toCharArray()) {            int tempNum = map.getOrDefault(ch, 0) + 1;            map.put(ch, tempNum);            maxFreq = Math.max(tempNum, maxFreq);        }        StringBuffer[] buckets = new StringBuffer[maxFreq + 1];        for (int i = 0; i <= maxFreq; i++) {            buckets[i] = new StringBuffer();        }
        for (Map.Entry<Character, Integer> entry : map.entrySet()) {            char c = entry.getKey();            int frequency = entry.getValue();            buckets[frequency].append(c);        }
        StringBuffer ans = new StringBuffer();        for (int i = maxFreq; i > 0; i--) {            StringBuffer bucket = buckets[i];            int size = bucket.length();            for (int j = 0; j < size; j++) {                for (int k = 0; k < i; k++) {                    ans.append(bucket.charAt(j));                }            }        }                return ans.toString();    }