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Given an array of integers and a number k, the majority number is the number that occurs more than 1/k of the size of the array. Find it.
There is only one majority number in the array.
For [3,1,2,3,2,3,3,4,4,4] and k = 3, return 3
O(n) time and O(k) extra space
Analysis:
Keep k-1 pointers and counters. Use HashMap to store numbers and counters, every delete operation will cost k-1, however, the max times of deletion is n/(k-1) because deletion only happens when there are (k-1) numbers in the map. So the overall complexity is still O(n).
Solution:
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 if (nums==null || nums.size()==0) return 0; 9 10 int len = nums.size(); 11 Map<Integer,Integer> nMap = new HashMap<Integer,Integer>(); 12 nMap.put(nums.get(0),1); 13 for (int i=1;i<len;i++) 14 if (nMap.containsKey(nums.get(i))){ 15 int val = nMap.get(nums.get(i))+1; 16 if (val==0) nMap.remove(nums.get(i)); 17 else nMap.put(nums.get(i),val); 18 } else { 19 //if the number of existing numbers is less than k-1, then just add one. 20 if (nMap.size()<k-1){ 21 nMap.put(nums.get(i),1); 22 } else { 23 List<Integer> keyList = new ArrayList<Integer>(); 24 //decrease the value of every key by 1. 25 for (Map.Entry en : nMap.entrySet()){ 26 int key = (int) en.getKey(); 27 int value = (int) en.getValue(); 28 value--; 29 if (value==0) keyList.add(key); 30 nMap.put(key,value); 31 } 32 for (int key : keyList) nMap.remove(key); 33 } 34 } 35 36 for (Map.Entry en: nMap.entrySet()) en.setValue(0); 37 int num = 0, count = 0; 38 for (int i=0;i<len;i++) 39 if (nMap.containsKey(nums.get(i))){ 40 int val = nMap.get(nums.get(i))+1; 41 if (val>count){ 42 num = nums.get(i); 43 count = val; 44 } 45 nMap.put(nums.get(i),val); 46 } 47 return num; 48 49 } 50 }
NOTE: It is a very good problem to practice HashMap and its iteration.
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原文地址：http://www.cnblogs.com/lishiblog/p/4189468.html
