Leetcode 451 Sort Characters By Frequency

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.

 思路比较简单，先统计再排序。
注意有可能出现相同频率的，所以说用一个list组成的array来进行排序。

public class Solution {
public String frequencySort(String s) {
if(s == null ){
return null;
}
if(s.length() == 0){
return "";
}
Map<Character, Integer> map = new HashMap<>();
int max = 0;
for(int i = 0; i < s.length(); i++){
char c = s.charAt(i);
if(map.containsKey(c)){
map.put(c, map.get(c) + 1);
}else{
map.put(c, 1);
}

max = Math.max(max, map.get(c) );
}
List<Character>[] list = new List[max + 1];
for (Character c : map.keySet()) {
int count = map.get(c);
if(list[count] == null){
list[count] = new ArrayList();
}
list[count].add(c);
}

StringBuilder sb = new StringBuilder();
for(int i = list.length - 1; i >= 0; i--){
if(list[i] != null){
for(Character c : list[i]){
for(int j = 0; j < i; j++){
sb.append(c);
}
}
}
}
return sb.toString();
}
}