LeetCode 229. Majority Element II

原题链接在这里：https://leetcode.com/problems/majority-element-ii/

题目：

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

Note: The algorithm should run in linear time and in O(1) space.

Example 1:

Input: [3,2,3]
Output: [3]

Example 2:

Input: [1,1,1,3,3,2,2,2]
Output: [1,2]

题解：

因为个数超过n/3的值最多有两个, 现在改成维护两个candidate.

最后再扫一遍数组，查看这两个最常出现值的实际出现次数，看看是否大于n/3, 大于的都加入返回res中.

Note: two candidate can't be equal, thus before adding candidate2, need to check if it is equal to candidate1, if yes, then skip.

Time Complexity: O(n). n = nums.length.

Space O(1).

AC Java:

class Solution {
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> res = new ArrayList<Integer>();
        if(nums == null || nums.length == 0){
            return res;
        }
        
        //最多有两个数字大于n/3
        int major1 = 1;
        int major2 = 2;
        int count1 = 0;
        int count2 = 0;
        for(int i = 0; i<nums.length; i++){
            if(nums[i] == major1){
                count1++;
            }else if(nums[i] == major2){
                count2++;
            }else if(count1 == 0){
                major1 = nums[i];
                count1 = 1;
            }else if(count2 == 0){
                major2 = nums[i];
                count2 = 1;
            }else{
                count1--;
                count2--;
            }
        }
        
        count1 = 0;
        count2 = 0;
        for(int num : nums){
            if(num == major1){
                count1++;
            }
            if(num == major2){
                count2++;
            }
        }
        
        if(count1 > nums.length/3){
            res.add(major1);
        }
        if(count2 > nums.length/3){
            res.add(major2);
        }
        return res;
    }
}

类似Majority Element I.