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; } };