229. Majority Element II

最新推荐文章于 2023-10-05 14:35:54 发布

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最新推荐文章于 2023-10-05 14:35:54 发布

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分类专栏： LeetCode Array 文章标签： LeetCode Array Majority Element II

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本文介绍了一个线性时间和O(1)空间复杂度的算法，用于找出数组中出现次数超过n/3的元素。该算法基于Boyer-Moore Majority Vote算法思想，并通过两次遍历实现。

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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.

基于Boyer-Moore Majority Vote algorithm的思想，代码如下：

public class Solution {
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> res = new ArrayList<Integer>();
        if (nums.length == 0) {
            return res;
        }
        int count1 = 0, count2 = 0, num1 = 0, num2 = 0;
        for (int num: nums) {
            if (num == num1) {
                count1 ++;
            } else if (num == num2) {
                count2 ++;
            } else if (count1 == 0) {
                num1 = num;
                count1 = 1;
            } else if (count2 == 0) {
                num2 = num;
                count2 = 1;
            } else {
                count1 --;
                count2 --;
            }
        }
        count1 = 0; count2 = 0;
        for (int num : nums) {
            if (num == num1) {
                count1 ++;
            } else if (num == num2) {
                count2 ++;
            }
        }
        if (count1 > nums.length / 3) {
            res.add(num1);
        }
        if (count2 > nums.length / 3) {
            res.add(num2);
        }
        return res;
    }
}