标签：tor app get class stringbu character compare cte base
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.
1 public class Solution { 2 public String frequencySort(String s) { 3 HashMap<Character, Integer> charFreqMap = new HashMap<>(); 4 for (int i = 0; i < s.length(); i++) { 5 char c = s.charAt(i); 6 charFreqMap.put(c, charFreqMap.getOrDefault(c, 0) + 1); 7 } 8 ArrayList<Map.Entry<Character, Integer>> list = new ArrayList<>(charFreqMap.entrySet()); 9 10 list.sort(new Comparator<Map.Entry<Character, Integer>>() { 11 public int compare(Map.Entry<Character, Integer> o1, Map.Entry<Character, Integer> o2) { 12 return o2.getValue().compareTo(o1.getValue()); 13 } 14 }); 15 StringBuffer sb = new StringBuffer(); 16 for (Map.Entry<Character, Integer> e : list) { 17 for (int i = 0; i < e.getValue(); i++) { 18 sb.append(e.getKey()); 19 } 20 } 21 return sb.toString(); 22 } 23 }
标签：tor app get class stringbu character compare cte base
原文地址：http://www.cnblogs.com/beiyeqingteng/p/6082124.html
