229. 求众数 II

给定一个大小为n的整数数组，找出其中所有出现超过⌊ n/3 ⌋次的元素。

进阶：尝试设计时间复杂度为O(n)、空间复杂度为O(1)的算法解决此问题。

示例 1：

输入：[3,2,3]
输出：[3]

示例 2：

输入：nums = [1]
输出：[1]

示例 3：

输入：[1,1,1,3,3,2,2,2]
输出：[1,2]

提示：

• 1 <= nums.length <= 5*10^4
• -10^9 <= nums[i] <= 10^9

1.暴力字典
Python解答：

class Solution:
    def majorityElement(self, nums: List[int]) -> List[int]:
        adic = {}
        for num in nums:
            if num not in adic.keys():
                adic[num] = 1
            else:
                adic[num] += 1
        return [key for key, value in adic.items() if value > len(nums)//3]

2.摩尔投票

class Solution:
    def majorityElement(self, nums: List[int]) -> List[int]:
        value = [None, None]
        count = [0, 0]
        i = 0
        while i < len(nums):
            if nums[i] == value[0]:
                count[0] += 1
            elif nums[i] == value[1]:
                count[1] += 1
            elif count[0] == 0:
                value[0] = nums[i]
                count[0] = 1
            elif count[1] == 0:
                value[1] = nums[i]
                count[1] = 1
            else:
                count[0] -= 1
                count[1] -= 1
            i += 1

        t = [0, 0]
        for i in range(len(nums)):
            if nums[i] == value[0]:
                t[0] += 1
            if nums[i] == value[1]:
                t[1] += 1
        return [value[i] for i in range(2) if t[i] > len(nums)//3]