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;

    }

};

All LeetCode Questions List 题目汇总

巴特西

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

All LeetCode Questions List 题目汇总

最新文章

热门文章