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

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.

 

给一个字符串按照字符出现的频率来排序。

Java:

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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:

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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:

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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++:

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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;     } };

　　

　　

 

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