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

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本文介绍了一种线性时间和常数空间复杂度的算法，用于在整数数组中找到出现次数超过n/3的所有元素。利用改进的摩尔投票算法，通过两次遍历找出可能的超频元素，并验证其是否满足条件。

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Question

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.

hint

How many majority elements could it possibly have?
Do you have a better hint?

THOUGHT

Moore’s Majority Vote
algorithm，The basic idea is based on Moore’s Voting Algorithm, we need two candidates with top 2 frequency. If meeting different number from the candidate, then decrease 1 from its count, or increase 1 on the opposite condition. Once count equals 0, then switch the candidate to the current number. The trick is that we need to count again for the two candidates after the first loop. Finally, output the numbers appearing more than n/3 times.

多数投票算法进行改进的。观察可知，数组中至多可能会有2个出现次数超过 ⌊ n/3 ⌋ 的众数记变量n1, n2为候选众数； c1, c2为它们对应的出现次数，遍历数组，记当前数字为num。
若num与n1或n2相同，则将其对应的出现次数加1；
否则，若c1或c2为0，则将其置为1，对应的候选众数置为num；
否则，将c1与c2分别减1；
最后，再统计一次候选众数在数组中出现的次数，若满足要求，则返回之。

CODE

public class Solution {
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> res = new ArrayList<Integer>();
        if(nums == null || nums.length == 0)
            return res;
        if(nums.length == 1){
            res.add(nums[0]);
            return res;
        }
        int majNum1 = nums[0],majNum2 = nums[0],count1 = 1,count2 = 0;
        for(int i = 0;i < nums.length;i++){
            if(nums[i] == majNum1)
                count1++;
            else if(nums[i] == majNum2)
                count2++;
            else if(count1 == 0){
                majNum1 = nums[i];
                count1 = 1;
            }

            else if(count2 == 0){
                majNum2 = nums[i];
                count2 = 1;
            }
            else{
                count1--;
                count2--;
            }
        }
        count1 = 0;
        count2 = 0;
        for(int i = 0;i < nums.length;i++){
            if(nums[i] == majNum1)
                count1++;
            else if(nums[i] == majNum2)
                count2++;
        }
        if(count1 > (nums.length/3))
            res.add(majNum1);
        if(count2 > (nums.length/3))
            res.add(majNum2);
        return res;
    }
}