There are duplicate elements

Q: Give you an array of integers nums . If any value appears in the array At least twice , return true ; If each element in the array is different from each other , return false .

Example

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

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

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

        Two ways of thinking , One is to use the method of sorting , Time complexity O(nlogn); The second is hash table , Time complexity O(N)
As an algorithm beginner , We are not just satisfied with “ adopt ”, It's comparing ideas , Extract its essence , Go to its bad place . Learn from others' strong points and close the gap , In this way, we can improve ourselves . The problem is not just to solve , We should also solve the problem in an efficient way , Only our way of thinking has changed , We will improve .
1. Sort

bool containsDuplicate(vector<int>& nums) {
     The idea of sorting   Time complexity O（nlogn）
	sort(nums.begin(),nums.end());
	if(nums.size()<2)// There is only one element or no element 
		return false;
	for(int i=0;i<=nums.size()-2;i++)// Other situations 
		if(nums[i]==nums[i+1])
			return false;
	return true;
}

2. Hashtable

bool containsDuplicate_plus(vector<int>& nums) {
    //  Using a hash table  O(n)
	unordered_set<int>s;
	for(auto x:nums){
    
		if(s.count(x)==0)
			s.insert(x);
		else
			return true;
	}
	return false;
}

The latter is a way of exchanging space for time .