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;

    }