原来C++有已经写好的哈希集合，

柳神曾经提到过可以过超时的unordered_set

 支持 auto迭代器

可以

insert

count

erase

size

clear

empty

#include <unordered_set>                // 0. include the library

int main() {
    // 1. initialize a hash set
    unordered_set<int> hashset;   
    // 2. insert a new key
    hashset.insert(3);
    hashset.insert(2);
    hashset.insert(1);
    // 3. delete a key
    hashset.erase(2);
    // 4. check if the key is in the hash set
    if (hashset.count(2) <= 0) {
        cout << "Key 2 is not in the hash set." << endl;
    }
    // 5. get the size of the hash set
    cout << "The size of hash set is: " << hashset.size() << endl; 
    // 6. iterate the hash set
    for (auto it = hashset.begin(); it != hashset.end(); ++it) {
        cout << (*it) << " ";
    }
    cout << "are in the hash set." << endl;
    // 7. clear the hash set
    hashset.clear();
    // 8. check if the hash set is empty
    if (hashset.empty()) {
        cout << "hash set is empty now!" << endl;
    }
}

https://leetcode-cn.com/explore/learn/card/hash-table/204/practical-application-hash-set/805/配套题目

存在重复元素

给定一个整数数组，判断是否存在重复元素。

如果任何值在数组中出现至少两次，函数返回 true。如果数组中每个元素都不相同，则返回 false。

示例 1:

输入: [1,2,3,1]
输出: true

示例 2:

输入: [1,2,3,4]
输出: false

示例 3:

输入: [1,1,1,3,3,4,3,2,4,2]
输出: true
class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        unordered_set<int> s;
        for(int i=0;i<nums.size();i++){
            if(s.count(nums[i])>0){
                return true;
            }s.insert(nums[i]);
        }return false;
    }
};