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

對於1/k 的數，每當要刪除的時候，要把k的數字同時刪掉，這樣不會影響最終的結果。 

 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 }