LeetCode Majority Element II(Moore Voting Algorithm即Majority Voting Algorithm)

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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.

题意：给出一个数组，求出现次数大于n/3的元素，要求时间复杂度为O(n),空间复杂度为O(1)

思路：用Majority Voting Algorithm的一般算法

1、候选者的个数为2,计数数组的元素个数也为2

2、在遍历时，如果候选者中没有包含，就将其插入，如果包含，将计数加1，如果没有包含，将所有候选者的计数减1

3、第二次遍历时，统计个数，判断出现次数是否满足要求

代码如下：

class Solution{    public List<Integer> majorityElement(int[] nums)    {        List<Integer> res = new ArrayList<Integer>();        if (nums.length == 0) return res;        else return __majorityElement(nums, 3);    }    private List<Integer> __majorityElement(int[] nums, int k)    {        int cnt = k - 1;        ArrayList<Integer> candidates = new ArrayList<Integer>();        ArrayList<Integer> count = new ArrayList<Integer>();        for (int i = 0; i < cnt; i++)        {            candidates.add(0);            count.add(0);        }        for (int num : nums)        {            boolean found = false;            for (int i = 0; i < cnt; i++)            {                if (count.get(i) == 0 || candidates.get(i) == num)                {                    int c = count.get(i);                    count.set(i,c + 1);                    candidates.set(i, num);                    found = true;                    break;                }            }            if (!found)            {                for (int i = 0; i < cnt; i++)                {                    int c = count.get(i);                    count.set(i, c - 1);                }            }        }        Collections.fill(count, 0);        for (int num : nums)        {            for (int i = 0; i < cnt; i++)            {                if (candidates.get(i) == num)                {                    int c = count.get(i);                    count.set(i, c + 1);                    break;                }            }        }        List<Integer> ans = new ArrayList<Integer>();        for (int i = 0; i < cnt; i++)        {            if (count.get(i) > nums.length / k)            {                ans.add(candidates.get(i));            }        }        return ans;    }}

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