[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:

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:

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:

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++:

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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