There are duplicate elements (deduplication, sorting)

Ritko Link: Rikko

        You are given an integer array nums. Returns true if any value occurs at least twice in the array; returns false if every element in the array is distinct.

Example 1:

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

Example 2:

input: nums = [1,2,3,4]
output: false

Example 3:

Input: nums = [1,1,1,3,3,4,3,2,4,2]
Output: true
 

hint:

1 <= nums.length <= 105
-109 <= nums[i] <= 109

Sample code 1: [set deduplication]

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        if len(nums) == len(set(nums)):
            return False
        else:
            return True

Idea: Use set to deduplicate. If there are duplicate elements, the length of the list must be smaller than the length of the original list after deduplication, so it is enough to judge whether the lengths of the two are the same

Sample code 2: [Sorting]

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        nums.sort()
        j = 1
        while j < len(nums):
            if nums[j] == nums[j-1]:
                return True
            else:
                j += 1
        return False

Idea: Sort first, then judge whether two adjacent elements are equal, if they are equal, it means that there are duplicate elements

Note: code 2 is more time consuming than code !