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]难度：medium题目：给定一整数数组，找出其元素出现次数大于其三分之一长度的所有元素。注意：算法时间复杂度O(n),空间复杂度为O(1)思路：投票法。1 找出三个不同的候选人，各消除一票，如果票数为0 退出候选人队列。2 最后对剩下的候选人重新统计票数Runtime: 2 ms, faster than 97.44% of Java online submissions for Majority Element II.Memory Usage: 39.8 MB, less than 29.68% of Java online submissions for Majority Element II.class Solution { public List<Integer> majorityElement(int[] nums) { List<Integer> result = new ArrayList<>(); if (null == nums) { return result; } int[] elems = new int[2]; int[] count = new int[2]; for (int i = 0; i < nums.length; i++) { if (elems[0] == nums[i]) { count[0]++; } else if (elems[1] == nums[i]) { count[1]++; } else { if (count[0] > 0 && count[1] > 0) { count[0]–; count[1]–; } else if (0 == count[0]) { elems[0] = nums[i]; count[0] = 1; } else if (0 == count[1]) { elems[1] = nums[i]; count[1] = 1; } } } int c0 = 0, c1 = 0; for (int i = 0; i < nums.length; i++) { if (count[0] > 0 && nums[i] == elems[0]) { c0++; } if (count[1] > 0 && nums[i] == elems[1]) { c1++; } } if (c0 > nums.length / 3) { result.add(elems[0]); } if (c1 > nums.length / 3) { result.add(elems[1]); } return result; }}