LeetCode Sort Characters By Frequency

原题链接在这里：https://leetcode.com/problems/sort-characters-by-frequency/

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Top K Frequent Elements. 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<Character, Integer> freqMap = new HashMap<Character, Integer>();
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<freq; i++){
19                 bucket[freq].append(c);
20             }
21         }
22
23         StringBuilder res = new StringBuilder();
24         for(int i = bucket.length-1; i>=0; i--){
25             if(bucket[i] != null){
26                 res.append(bucket[i]);
27             }
28         }
29         return res.toString();
30     }
31 }