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) + 1);
            } else {
                map.put(num, 1);
            }
        }
        
        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 = 0; 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