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169. Majority Element

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

Example 1:

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

Example 2:

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

解题思路

（1）方法一：快排（Time Limit Exceeded）

    # 方法一： 快排
    def majorityElement(self, nums: List[int]) -> int:
        # 快排
        nums = self.quickSort(nums, 0, len(nums) - 1)
        # 直接返回中位值
        return nums[len(nums) // 2]

    def quickSort(self, nums, low, high):
        if low < high:
            index = self.partition(nums, low, high)
            self.quickSort(nums, low, index - 1)
            self.quickSort(nums, index + 1, high)
        return nums

    def partition(self, nums, low, high):
        key = nums[low]
        while low < high:
            while low < high and nums[high] >= key:
                high -= 1
            nums[low] = nums[high]

            while low < high and nums[low] <= key:
                low += 1
            nums[high] = nums[low]

        nums[high] = key

        return high

（2）方法二：基于Partition函数的时间复杂度为O(n)算法（Time Limit Exceeded）

class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        middle = len(nums) // 2
        index = self.partition(nums, 0, len(nums) - 1)
        while index != middle:
            if index > middle:
                index = self.partition(nums, 0, index - 1)
            else:
                index = self.partition(nums, index + 1, len(nums) - 1)

        return nums[middle]

    def partition(self, nums, low, high):
        key = nums[low]
        while low < high:
            while low < high and nums[high] >= key:
                high -= 1
            nums[low] = nums[high]

            while low < high and nums[low] < key:
                low += 1
            nums[high] = nums[low]

        nums[high] = key

        return high

（3）方法三：用字典统计（Accepted）

    # 方法三：用字典统计
    def majorityElement3(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return nums[0]
        dic = {}
        for i in nums:
            if i in dic.keys():
                dic[i] += 1
                if dic[i] > (len(nums) // 2):
                    return i
            else:
                dic[i] = 1