LeetCode 451. Sort Characters By Frequency

1. Priority Queue
   时间复杂度应该是O(n + k*lgk)，n是输入string的长度，k是string里面unique characters的数量。

   class Solution {
       public String frequencySort(String s) {
           Map<Character, Integer> count = new HashMap<>();
           for(char c: s.toCharArray()){
               count.put(c, count.getOrDefault(c, 0) + 1);
           }
           // Our comparator is (a, b) -> b.length() - a.length().
           // If a is longer than b, then a negative number will be returned
           // telling the sort algorithm to place a first. Otherwise, a positive
           // number will be returned, telling it to place a second.
           // This results in a longest to shortest sorted list of the strings.
           Queue<Character> pq = new PriorityQueue<>((a,b) 
           -> count.get(b) - count.get(a));
           // The keySet() method is used to get a Set view of the keys 
           // contained in this map.
           for(char c: count.keySet()){
               pq.offer(c);
           }
           StringBuilder sb = new StringBuilder();
           while(!pq.isEmpty()){
               char c = pq.poll();
               int copies = count.get(c);
               for(int i = 1; i <= copies; i++){
                   sb.append(c);
               }
           }
           return sb.toString();
       }
   }
   

    

2. Bucket Sort
   我们用一个maxFreq变量来维护频率的最大值，这样我们只需要建立(maxFreq + 1)个bucket就行了，时间复杂度是O(n）。
   第一种写法：建立List<List<Character>>来承担buckets的作用

   class Solution {
       public String frequencySort(String s) {
           Map<Character, Integer> count = new HashMap<>();
           int maxFreq = 0;
           // Count up the occurances.
           for(char c: s.toCharArray()){
               count.put(c, count.getOrDefault(c, 0) + 1);
               maxFreq = Math.max(maxFreq, count.get(c));
           }
           // Make the list of buckets and apply bucket sort.
           List<List<Character>> buckets = new ArrayList<>();
           // We would like the freqency to become the indices
           // i.e. indices range from 0 to maxFreq
           // hence we need to create a list whose size is maxFreq + 1
           for(int i = 0; i <= maxFreq; i++){
               List<Character> bucket = new ArrayList<>();
               buckets.add(bucket);
           }
           // Add characters to different buckets acoording to their occurences
           for(char c: count.keySet()){
               int freq = count.get(c);
               buckets.get(freq).add(c);
           }
           // Build up the string
           StringBuilder sb = new StringBuilder();
           for(int i = buckets.size() - 1; i >= 1; i--){
               for(char c: buckets.get(i)){
                   for(int j = count.get(c); j >= 1; j--){
                       sb.append(c);
                   }
               }
           }
           return sb.toString();
       }
   }

   第二种写法：创建一个元素为List[]的array。但是有两点要注意，一是创建List[]的语法格式，https://blog.youkuaiyun.com/huanghanqian/article/details/81078799 ； 二是从后往前遍历这个数组的时候（18~26）需要check数组的元素是否为空。因为之前只有往数组里面加char时，我们才会创建对应frequency的ArrayList，否则是null。这是和第一种法所不一样的，第一种写法我们一开始就创建好了所有的List（15~18）。
    

   class Solution {
       public String frequencySort(String s) {
           Map<Character, Integer> count = new HashMap<>();
           int maxFreq = 0;
           for(char c: s.toCharArray()){
               count.put(c, count.getOrDefault(c, 0) + 1);
               maxFreq = Math.max(maxFreq, count.get(c));
           }
           List<Character>[] buckets = new List[maxFreq + 1];
           for(char c: count.keySet()){
               int freq = count.get(c);
               if(buckets[freq] == null){
                   buckets[freq] = new ArrayList<Character>();
               }
               buckets[freq].add(c);
           }
           StringBuilder sb = new StringBuilder();
           for(int i = buckets.length - 1; i > 0; i--){
               if(buckets[i] != null){
                   for(char c: buckets[i]){
                       for(int j = 1; j <= count.get(c); j++){
                           sb.append(c);
                       }
                   }
               } 
           }
           return sb.toString();
       }
   }