Majority Element II——LeetCode

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.

 

题目大意：给定一个大小为n的数组，找出Majority元素，满足出现次数大于 ⌊ n/3 ⌋ 次。

解题思路：大于 ⌊ n/3 ⌋ 次，那么应该最多有两个Majority元素，之前有道题是找出 大于⌊ n/2 ⌋ 次Majority元素，这道题有点类似，稍微复杂一点，还是利用投票算法，count作为对元素的计数，候选元素等于当前元素则count+1，否则减1，count等于0则更换候选元素。

    public List<Integer> majorityElement(int[] nums) {
        List<Integer> res = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return res;
        }
        int can1 = nums[0], can2 = nums[0], cnt1 = 1, cnt2 = 0;
        for (int i = 1; i < nums.length; i++) {
            int num = nums[i];
            if ((cnt1 == 0 && num != can2)) {
                can1 = num;
            } else if (cnt2 == 0 && num != can1) {
                can2 = num;
            }
            if (num == can1) {
                cnt1++;
            } else if (num == can2) {
                cnt2++;
            } else {
                cnt1--;
                cnt2--;
            }
            cnt1 = cnt1 < 0 ? 0 : cnt1;
            cnt2 = cnt2 < 0 ? 0 : cnt2;
        }
        cnt1 = 0;
        cnt2 = 0;
        for (int num : nums) {
            if (num == can1) {
                cnt1++;
            } else if (num == can2) {
                cnt2++;
            }
        }
        if (cnt1 > nums.length / 3)
            res.add(can1);
        if (cnt2 > nums.length / 3)
            res.add(can2);
        return res;
    }