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
My solution need to sort the list first which will be O(nlogn) but won’t have any extra space
class Solution:
def majorityElement(self, nums: List[int]) -> int:
nums = sorted(nums)
return nums[int(len(nums)/2)]
It can be also done by HashMap.
The magic is Moore Voting Algorithm
class Solution:
def majorityElement(self, nums: List[int]) -> int:
candidate = None
count = 0
for n in nums:
if count == 0:
candidate = n
count += 1
elif n == candidate:
count += 1
else:
count -= 1
return candidate
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