Easy
Given an array nums of size n, return the majority element.
The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.
Example 1:
Input: nums = [3,2,3] Output: 3
Example 2:
Input: nums = [2,2,1,1,1,2,2] Output: 2
Constraints:
n == nums.length1 <= n <= 5 * 104-109 <= nums[i] <= 109
Follow-up: Could you solve the problem in linear time and in
O(1) space? class Solution:
def majorityElement(self, nums: List[int]) -> int:
# c = Counter(nums)
# return c.most_common()[0][0]
count = 0
candidate = None
for num in nums:
if count == 0:
candidate = num
count += (1 if num == candidate else -1)
return candidate