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
-231 <= nums[i] <= 231 - 1
[방법1] 정렬후 순차탐색하며 maxCount의 원소 찾기
[방법2] 해쉬맵을 이용하여 각 원소별 횟수 카운팅
[방법3] Boyer-Moore Voting Algorithm (솔루션 참고)
- majority element는 전체 배열의 2/n 개 이상을 차지함.
class Solution {
public int majorityElement(int[] nums) {
int count = 0;
int candidate = 0;
for (int num : nums) {
if (count == 0) {
candidate = num;
}
count += (num == candidate) ? 1 : -1;
}
return candidate;
}
}