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.

Solution1：Moore voting

和169題思路Solution1相同： http://www.lxweimin.com/p/931ed67158fe
思路：主要思想是每次找到三個不同的元素 一起棄掉，最終剩下的就是有可能是是出現過>n/3的。
最多只能有兩個出現過>n/3次的，所以定義兩個candidates，如果有的話，Moore voting會選出來，但題目沒有guarantee，所以最后需要再遍歷count來double check。
Time Complexity: O(N) Space Complexity: O(1)

Solution2：二分治來找到最多出現的兩個 再check？

Solution3：Sort

思路：sort后 找到 1/3處 和 2/3處 這兩個candidates 。再double check 是否超過n/3次，check的方式因為已經排好序，可以binary search 找 起始位置和終止位置，check類似81題：http://www.lxweimin.com/p/597d80b0c161
Solution3 refer: https://discuss.leetcode.com/topic/65330/a-binary-search-solution-assuming-the-input-array-is-sorted-beats-99/2

Code: not finished

Solution4：Hashmap

Solution1 Code：

class Solution1 {
    // Moore voting algorithm
    public List<Integer> majorityElement(int[] nums) {
        
        List<Integer> result = new ArrayList<Integer>();
        
        int candidate1 = 0, candidate2 = 0; // there are max two candidates
        int count1 = 0, count2 = 0;
        for(int i = 0; i < nums.length; i++) {
            if(nums[i] == candidate1) {
                count1++;
            } else if (nums[i] == candidate2) {
                count2++;
            } else if(count1 == 0) {
                candidate1 = nums[i];
                count1 = 1;
            } else if(count2 == 0) {
                candidate2 = nums[i];
                count2 = 1;
            } else {
                count1--;
                count2--;
            }
        }
        
        // check two candidiates, since there is no guarantee there are two that appear > n / 3
        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 > nums.length / 3)
            result.add(candidate1);
        if (count2 > nums.length / 3)
            result.add(candidate2);
        return result;
    }
        
}