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.

一刷
题解：类似于169. Majority Element。用两个count变量，和两个candidate
第一次iter:
如果nums[I] == candidate1, count1++;
如果nums[I] == candidate2, count2++;
如果count1 == 0, candidate1 = nums[I];
如果count2 == 0, candidate2 = nums[I];
else count1--, count2--;

第二次iter:
统计candidate1, candidate2出现次数，如果大于len/3，加入结果集中

public class Solution {
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> res = new ArrayList<>();
        if(nums == null || nums.length == 0) return res;
        
        int count1 = 0, count2 = 0, candidate1 = 0, candidate2 = 0;
        int len = nums.length;
        for(int num : nums){
            if(num == candidate1){
                count1 += 1;
            }
            else if(num == candidate2){
                count2 += 1;
            }
            else if(count1 == 0){
                candidate1 = num;
                count1 = 1;
            }
            else if(count2 == 0){
                candidate2 = num;
                count2 = 1;
            }
            else{
                count1--;
                count2--;
            }
        }
        count1 = 0;
        count2 = 0;
        for(int i=0; i<nums.length; i++){
            if(nums[i] == candidate1) count1++;
            else if(nums[i] == candidate2) count2++;
        }
        if(count1>len/3) res.add(candidate1);
        if(count2>len/3) res.add(candidate2);
        return res;
    }
}