Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.
Example 1:
Input:nums = [1,2,3,1]
Output:true
Explanation:
The element 1 occurs at the indices 0 and 3.
Example 2:
Input:nums = [1,2,3,4]
Output:false
Explanation:
All elements are distinct.
Example 3:
Input:nums = [1,1,1,3,3,4,3,2,4,2]
Output:true
Constraints:
1 <= nums.length <= 105
-109 <= nums[i] <= 109
Solutions
Solution 1: Sorting
First, we sort the array nums.
Then, we traverse the array. If there are two adjacent elements that are the same, it means that there are duplicate elements in the array, and we directly return true.
Otherwise, when the traversal ends, we return false.
The time complexity is $O(n \times \log n)$, where $n$ is the length of the array nums.
We traverse the array and record the elements that have appeared in the hash table $s$. If an element appears for the second time, it means that there are duplicate elements in the array, and we directly return true.
The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array nums.