# [LeetCode] 451. Sort Characters By Frequency 根据字符出现频率排序

2021年09月15日 阅读数：1

Given a string, sort it in decreasing order based on the frequency of characters.html

Example 1:java

```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:python

```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:app

```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.```

Java:url

```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();
}
}　```

Python:htm

```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())　```

Python:blog

```class Solution(object):
def frequencySort(self, s):
"""
:type s: str
:rtype: str
"""
freq = collections.defaultdict(int)
for c in s:
freq[c] += 1

counts = [""] * (len(s)+1)
for c in freq:
counts[freq[c]] += c

result = ""
for count in reversed(xrange(len(counts)-1)):
for c in counts[count]:
result += c * count

return result
```

C++:排序

```class Solution {
public:
string frequencySort(string s) {
unordered_map<char, int> freq;
for (const auto& c : s) {
++freq[c];
}

vector<string> counts(s.size() + 1);
for (const auto& kvp : freq) {
counts[kvp.second].push_back(kvp.first);
}

string result;
for (int count = counts.size() - 1; count >= 0; --count) {
for (const auto& c : counts[count]) {
result += string(count, c);
}
}

return result;
}
};
```