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.
解题报告
使用桶排序法，第n个桶里有出现频率为n的字符。再从后面往前遍历，不断加入为n的字符，返回这个值。
class Solution {
public String frequencySort(String s) {
HashMap<Character, Integer> map = new HashMap();
for(char ss : s.toCharArray()){
map.put(ss, map.getOrDefault(ss, 0) + 1);
}
// create a array which can store list
List<Character>[] bucket = new ArrayList[s.length() + 1];
for(char c : map.keySet()){
int val = map.get(c);
if(bucket[val] == null){
bucket[val] = new ArrayList();
}
bucket[val].add(c);
}
StringBuilder sb = new StringBuilder();
// run through from larger freq to lower freq
for(int i = bucket.length - 1; i >=0; i--){
if(bucket[i] == null)
continue;
for(char c : bucket[i]){
for(int j = i; j > 0 ; j--)
sb.append(c);
}
}
return sb.toString();
}
}