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
Follow-up: Could you solve the problem in linear time and in
O(1)
space?
Companies:
Microsoft, Apple, Facebook, Amazon, Google, Adobe, Rubrik
Related Topics:
Array, Hash Table, Divide and Conquer, Sorting, Counting
Similar Questions:
// OJ: https://leetcode.com/problems/majority-element/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
int majorityElement(vector<int>& nums) {
int ans = 0, cnt = 0;
for (int n : nums) {
if (ans == n) ++cnt;
else if (cnt > 0) --cnt;
else {
ans = n;
cnt = 1;
}
}
return ans;
}
};