[刷题]Majority Number III
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[LintCode]Majority Number III
Version 1
public class Solution { /** * @param nums: A list of integers * @param k: As described * @return: The majority number */ public int majorityNumber(ArrayList<Integer> nums, int k) { // 2015-09-22 O(n) if (nums == null || nums.size() == 0 || k < 1) { return 0; } HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.size(); i++) { int temp = nums.get(i); if (!map.containsKey(temp)) { map.put(temp, 1); } else { map.put(temp, map.get(temp) + 1); } } int len = nums.size(); for (Integer i : map.keySet()) { if (map.get(i) > len / k) { return i; } } return 0; }}
Version 2
public class Solution { /** * @param nums: A list of integers * @param k: As described * @return: The majority number */ public int majorityNumber(ArrayList<Integer> nums, int k) { // O(n) time and O(k) extra space // count at most k keys. HashMap<Integer, Integer> counters = new HashMap<Integer, Integer>(); for (Integer i : nums) { if (!counters.containsKey(i)) { counters.put(i, 1); } else { counters.put(i, counters.get(i) + 1); } if (counters.size() >= k) { removeKey(counters); } } // corner case if (counters.size() == 0) { return Integer.MIN_VALUE; } // recalculate counters for (Integer i : counters.keySet()) { counters.put(i, 0); } for (Integer i : nums) { if (counters.containsKey(i)) { counters.put(i, counters.get(i) + 1); } } // find the max key int maxCounter = 0, maxKey = 0; for (Integer i : counters.keySet()) { if (counters.get(i) > maxCounter) { maxCounter = counters.get(i); maxKey = i; } } return maxKey; } /** * 不能在遍历的时候删除元素,所以在遍历结束后再删除 */ private void removeKey(HashMap<Integer, Integer> counters) { Set<Integer> keySet = counters.keySet(); List<Integer> removeList = new ArrayList<>(); for (Integer key : keySet) { counters.put(key, counters.get(key) - 1); if (counters.get(key) == 0) { removeList.add(key); } } for (Integer key : removeList) { counters.remove(key); } }}
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