# 451 Sort Characters By Frequency

### Problem:

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.
``````

### Solutions:

``````public class Solution {
public String frequencySort(String s) {
HashMap<Character, Integer> appr = new HashMap<Character, Integer>();
int max = Integer.MIN_VALUE;
for (int i = 0; i < s.length(); i ++) {
char c = s.charAt(i);
if (!appr.containsKey(c)) {
appr.put(c, 0);
}
appr.put(c, appr.get(c) + 1);
max = Math.max(max, appr.get(c));
}
HashMap<Integer, Queue<Character>> index = new HashMap<Integer, Queue<Character>>();
for (Character c:appr.keySet()) {
int count = appr.get(c);
if (!index.containsKey(count)) {
index.put(count, new LinkedList<Character>());
}
index.get(count).add(c);
}
StringBuilder sb = new StringBuilder();
for (int i = max; i > 0; i --) {
if (index.containsKey(i)) {
while (!index.get(i).isEmpty()) {
char c = index.get(i).poll();
for (int j = 0; j < i; j ++) {
sb.append(c);
}
}
}

}
return sb.toString();
}
}
``````