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.length
1 <= n <= 5 * 104
-109 <= nums[i] <= 109
Follow-up: Could you solve the problem in linear time and in O(1)
space?
-spec majority_element(Nums :: [integer()]) -> integer().
majority_element(Nums) ->
element(1, lists:foldl(
fun(Num, {Candi, Cnt}) ->
NewCandi = if Cnt == 0 -> Num; true -> Candi end,
NewCnt = if NewCandi == Num -> Cnt + 1; true -> Cnt - 1 end,
{NewCandi, NewCnt}
end,
{0, 0},
Nums
)).