leetcode__求众数__python

解题思路:

这道题总体来说还是比较简单，维持两个中间变量即可，根据出现的次数一定大于或者等于总次数的一半，那么我们可以得到代码如下：

class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        #keep a count
        #keep a key
        count = 0
        key = 'start'
        for num in nums:
            if count == 0 and key != num:
                key = num
                count += 1
            elif count == 0 and key == num:
                count += 1
            elif count > 0 and key == num:
                count += 1
            elif count > 0 and key != num:
                count -= 1
        return key