九章面试题：Find first K frequency numbers 解题报告

Find first K frequency numbers

/*
 * Input: int[] A = {1, 1, 2, 3, 4, 5, 2}; k = 3
 * return the highest frequency numbers.
 * return: [1, 2, 3] or [1, 2, 4] or [1, 2, 5]
 * */

找出出现频率前k的数字

SOLUTION 1:

先将数字全放入一个map, key为数字，value为frequency, 对map排序，时间复杂度是NlogN。注意使用comparator.

 /*
      * Solution 1:
      * 对HashMap Sort.
      * Complexity: O(NLogN)
      * */
     public static Set<Integer> findKthFrenquency1(int[] input, int k) {
         HashSet<Integer> set = new HashSet<Integer>();
         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
         for (int num: input) {
             if (map.containsKey(num)) {
                 map.put(num, map.get(num) + );
             } else {
                 map.put(num, );
             }
         }

         ArrayList<Entry<Integer, Integer>> list = new ArrayList<Entry<Integer, Integer>>(map.entrySet());

         Collections.sort(list, new Comparator<Entry<Integer, Integer>>() {
             public int compare(Entry<Integer, Integer> o1, Entry<Integer, Integer> o2) {
                 return o2.getValue() - o1.getValue();
             }
         });

         for (int i = ; i < k; i++) {
             set.add(list.get(i).getKey());
         }

         return set;
     }

SOLUTION 2:

使用TreeMap, 待补充

SOLUTION 3:

使用优先队列，建立一个k+1size的PriorityQueue,然后每次先拉出一个频率最小值，再添加一个值，全部完成后，频率最大的k个数字会留在队列中。

时间复杂度：NlogK, 如果K比较小的时候，就相当于N了。

 /*
      * Solution 3:
      * Use The priority queue.
      * */
     public static List<Integer> findKthFrenquency(int[] input, int k) {
         LinkedList<Integer> list = new LinkedList<Integer>();

         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
         for (int num: input) {
             if (map.containsKey(num)) {
                 map.put(num, map.get(num) + 1);
             } else {
                 map.put(num, 1);
             }
         }

         PriorityQueue<Entry<Integer, Integer>> q = new PriorityQueue<Entry<Integer, Integer>>(k + 1, new Comparator<Entry<Integer, Integer>>(){
            public int compare(Entry<Integer, Integer> o1, Entry<Integer, Integer> o2) {
                return o1.getValue() - o2.getValue();
            }
         });

         for (Entry<Integer, Integer> entry: map.entrySet()) {
             if (q.size() == k + 1) {
                 // Delete the smallest element from the queue.
                 q.poll();
             }
             q.offer(entry);
         }

         // delete one small element
         q.poll();

         while (!q.isEmpty()) {
             Entry<Integer, Integer> entry = q.poll();
             list.addFirst(entry.getKey());
         }

         return list;
     }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/algorithm/interviews/pocketGem/FindKthFrequency.java