Majority Number III

Given an array of integers and a number k, the majority number is the number that occursmore than 1/k of the size of the array.

Find it.

Example

Given [3,1,2,3,2,3,3,4,4,4] and k=3, return 3.

Note

There is only one majority number in the array.

Challenge

O(n) time and O(k) extra space

对于1/k 的数，每当要删除的时候，要把k的数字同时删掉，这样不会影响最终的结果。 

 1 public class Solution {

 2     /**

 3      * @param nums: A list of integers

 4      * @param k: As described

 5      * @return: The majority number

 6      */

 7     public int majorityNumber(ArrayList<Integer> nums, int k) {

 8         // write your code

 9         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();

10         for(int i = 0; i < nums.size(); i ++){

11             int num = nums.get(i);

12             if(map.containsKey(num)){

13                 map.put(num, map.get(num) + 1);

14             }else{

15                 if(map.size() < 3){

16                     map.put(num, 1);

17                 }else{//decrease every count by one, remove pair which count is 0

18                     ArrayList<Integer> removeKey = new ArrayList<Integer>();

19                     for (Map.Entry en : map.entrySet()){

20                         en.setValue((int)en.getValue() - 1);

21                         if((int)en.getValue() == 0){

22                             removeKey.add((int)en.getKey());

23                         }

24                     }

25                     for(int j = 0; j < removeKey.size(); j ++){

26                         map.remove(removeKey.get(j));

27                     }

28                 }

29             }

30         }

31         

32         int result = 0;

33         int resultValue = 0;

34         for (Map.Entry cur : map.entrySet()){

35             if((int)cur.getValue() > resultValue){

36                 result = (int)cur.getKey();

37                 resultValue = (int)cur.getValue();

38             }

39         }

40         return result;

41     }

42 }