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# 0229. Majority Element II

<https://leetcode.com/problems/majority-element-ii>

## Description

Given an integer array of size `n`, find all elements that appear more than `⌊ n/3 ⌋` times.

**Follow-up:** Could you solve the problem in linear time and in O(1) space?

**Example 1:**

```
**Input:** nums = [3,2,3]
**Output:** [3]
```

**Example 2:**

```
**Input:** nums = [1]
**Output:** [1]
```

**Example 3:**

```
**Input:** nums = [1,2]
**Output:** [1,2]
```

**Constraints:**

* `1 <= nums.length <= 5 * 104`
* `-109 <= nums[i] <= 109`

## ac1 Boyer-Moore Algorithm

```java
class Solution {
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> res = new ArrayList<Integer>();
        // corner case
        if (nums.length == 0) return res;

        // at most 2 majority. pass1: filter out majorities. pass2: recount numbers
        // pass1: filter out majorities
        int m1 = nums[0], m2 = nums[0], cnt1 = 0, cnt2 = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == m1) {
                cnt1++;
            } else if (nums[i] == m2) {
                cnt2++;
            } else if (cnt1 == 0) {
                m1 = nums[i];
                cnt1++;
            } else if (cnt2 == 0) {
                m2 = nums[i];
                cnt2++;
            } else {
                cnt1--;
                cnt2--;
            }
        }

        // pass2
        cnt1 = 0; cnt2 = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == m1) cnt1++;
            else if (nums[i] == m2) cnt2++;
        }

        // Determine
        if (cnt1 > nums.length/3) res.add(m1);
        if (cnt2 > nums.length/3) res.add(m2);

        return res;
    }
}
```


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