LeetCode-in-All
169. Majority Element
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?
Solution
func majorityElement(nums []int) int {
count := 1
majority := nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] == majority {
count++
} else {
if count > 1 {
count--
} else {
majority = nums[i]
}
}
}
count = 0
for _, num := range nums {
if num == majority {
count++
}
}
if count >= (len(nums)/2)+1 {
return majority
} else {
return -1
}
}