229. Majority Element II

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.

Note: The algorithm should run in linear time and in O(1) space.

Example 1:

Input: [3,2,3]
Output: [3]

Example 2:

Input: [1,1,1,3,3,2,2,2]
Output: [1,2]

class Solution {
    public List<Integer> majorityElement(int[] nums) {
            ArrayList<Integer> res = new ArrayList<Integer>();
            if (nums.length==0) return res;

            int count[] = new int[2];        
            int x[] = new int[2];       

            x[0] = 0; x[1] = 1;        
            for (int i = 0; i < nums.length; i++) {
                if (x[0] == nums[i])
                    count[0]++;
                else if (x[1] == nums[i])
                    count[1]++;
                else if (count[0] == 0) {
                    x[0] = nums[i];
                    count[0] = 1;
                } else if (count[1] == 0) {
                    x[1] = nums[i];
                    count[1] = 1;
                } else {
                    count[0]--;
                    count[1]--;                
                }
            }

            Arrays.fill(count, 0);
            for (int i : nums) {// Count again for x1, x2
            if (i == x[0]) count[0]++;
            else if (i == x[1]) count[1]++;
        }
        for (int j = 0; j < 2; j++) {
            if (count[j] > nums.length/3 && !res.contains(x[j])) res.add(x[j]);
        }        
        return res;
    }
}

https://leetcode.com/problems/majority-element-ii/discuss/63584/Concise-JAVA-solution-based-on-Moore's-Voting-Algorithm

还是boyer-moore投票法，这次要的是大于1/3，说明最多有两个candidate，initialize两个candidate，和1/2一样相等就++，小于等于0就换，不一样就--。

最后得到了两个candidates，重新计算两个candidates的counts，成为result