Problem

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:

1
2
3
Input: nums = [3,2,3]
Output: 3

Example 2:

1
2
3
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

1
O(1)

space?

Solve

解1

1
2
3
4
5
6
7
8
9
10
class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        dicts = {}
        for i in range(len(nums)):
            if nums[i] in dicts:
                dicts[nums[i]]+=1
            else:
                dicts[nums[i]] = 1
            if dicts[nums[i]] > len(nums) / 2:
                return nums[i]

簡化

1
2
3
4
5
6
7
8
9
10
11
class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        dicts = {}
        for num in nums:
            if num in dicts:
                dicts[num] += 1
            else:
                dicts[num] = 1

            if dicts[num] > len(nums) / 2:
                return num

特殊解法

1
2
3
4
5
6
7
8
9
10
11
class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        nums.sort()
        ans = len(nums) // 2
        return nums[ans]
        
class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        nums.sort()
        n = len(nums)
        return nums[n//2]