## 题目描述：

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.
```

## Python代码：

``````class Solution(object):
def frequencySort(self, s):
"""
:type s: str
:rtype: str
"""
return ''.join(c * t for c, t in collections.Counter(s).most_common())
``````

## Java代码：

``````public class Solution {
public String frequencySort(String s) {
HashMap<Character, Integer> charFreqMap = new HashMap<>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
charFreqMap.put(c, charFreqMap.getOrDefault(c, 0) + 1);
}
ArrayList<Map.Entry<Character, Integer>> list = new ArrayList<>(charFreqMap.entrySet());
list.sort(new Comparator<Map.Entry<Character, Integer>>(){
public int compare(Map.Entry<Character, Integer> o1, Map.Entry<Character, Integer> o2) {
return o2.getValue().compareTo(o1.getValue());
}
});
StringBuffer sb = new StringBuffer();
for (Map.Entry<Character, Integer> e : list) {
for (int i = 0; i < e.getValue(); i++) {
sb.append(e.getKey());
}
}
return sb.toString();
}
}
``````

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