Majority Element II-优快云博客

本文介绍了一个线性时间和O(1)空间复杂度的算法，用于找出长度为n的整数数组中出现次数超过n/3的所有元素。通过使用HashMap来跟踪潜在的多数元素，并最终验证这些候选是否确实满足条件。

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Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space.

public class Solution {
    public List<Integer> majorityElement(int[] nums) {
    	List<Integer> res = new ArrayList<>();
        printKMajor(nums, 3, res);
        return res;
    }

	private void printKMajor(int[] nums, int k, List<Integer> res) {
		// TODO Auto-generated method stub
		if (k < 2) {
			return;
		}
		HashMap<Integer,Integer> cands = new HashMap<>();
		for (int i = 0; i < nums.length; i++) {
			if (cands.containsKey(nums[i])) {
				cands.put(nums[i], cands.get(nums[i]) + 1);
			} else {
				if (cands.size() == k - 1) {
					allCandsNinusOne(cands);
				} else {
					cands.put(nums[i], 1);
				}
			}
		}
		HashMap<Integer,Integer> reals = getReals(nums, cands);
		for (Map.Entry<Integer, Integer> set : reals.entrySet()) {
			Integer key = set.getKey();
			if (reals.get(key) > nums.length / k) {
				res.add(key);
			}
		}
	}

	private HashMap<Integer,Integer> getReals(int[] nums, HashMap<Integer, Integer> cands) {
		// TODO Auto-generated method stub
		HashMap<Integer,Integer> hashMap = new HashMap<>();
		for (int i = 0; i < nums.length; i++) {
			int num = nums[i];
			if (cands.containsKey(num)) {
				if (hashMap.containsKey(num)) {
					hashMap.put(num, hashMap.get(num) + 1);
				} else {
					hashMap.put(num, 1);
				}
			}
		}
		return hashMap;
	}

	private void allCandsNinusOne(HashMap<Integer, Integer> map) {
		// TODO Auto-generated method stub
		LinkedList<Integer> linkedList = new LinkedList<>();
		for (Map.Entry<Integer, Integer> set : map.entrySet()) {
			Integer key = set.getKey();
			Integer value = set.getValue();
			if (value == 1) {
				linkedList.add(key);
			}
			map.put(key, value - 1);
		}
		for (Integer integer : linkedList) {
			map.remove(integer);
		}
	}
    
}