Contains Duplicate II Solutions in C++

// Source : https://leetcode.com/problems/contains-duplicate-ii/
// Author : Hao Chen
// Date   : 2015-06-12



class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {
        unordered_map<int, int> m;
        for (int i=0; i<nums.size(); i++) {
            int n = nums[i];
            if (m.find(n) != m.end() && i - m[n] <= k) {
                return true;
            }
            m[n] = i;
        }
        return false;
    }
};

/// Source : https://leetcode.com/problems/contains-duplicate-ii/description/
/// Author : liuyubobobo
/// Time   : 2017-11-15

#include <iostream>
#include <vector>
#include <unordered_set>

using namespace std;

/// Using Hash Set
/// Time Complexity: O(n)
/// Space Complexity: O(k)
class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {

        if(nums.size() <= 1)
            return false;

        if(k <= 0)
            return false;

        unordered_set<int> record;
        for(int i = 0 ; i < nums.size() ; i ++){

            if(record.find(nums[i]) != record.end())
                return true;

            record.insert(nums[i]);

            // 保持record中最多有k个元素
            // 因为在下一次循环中会添加一个新元素,使得总共考虑k+1个元素
            if(record.size() == k + 1)
                record.erase(nums[i - k]);
        }

        return false;
    }
};


void print_bool(bool b){
    cout << (b ? "True" : "False") << endl;
}

int main() {

    vector<int> nums = {1, 2, 1};
    int k = 1;

    print_bool(Solution().containsNearbyDuplicate(nums, k));

    return 0;
}

/// Source : https://leetcode.com/problems/contains-duplicate-ii/description/
/// Author : liuyubobobo
/// Time   : 2019-04-10

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

/// Using Hash Map
/// Time Complexity: O(n)
/// Space Complexity: O(n)
class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {

        if(nums.size() <= 1)
            return false;

        if(k <= 0)
            return false;

        unordered_map<int, int> record;
        record[nums[0]] = 0;
        for(int i = 1 ; i < nums.size() ; i ++){

            if(record.count(nums[i]) && i - record[nums[i]] <= k)
                return true;

            record[nums[i]] = i;
        }

        return false;
    }
};


void print_bool(bool b){
    cout << (b ? "True" : "False") << endl;
}

int main() {

    vector<int> nums = {1, 2, 1};
    int k = 1;

    print_bool(Solution().containsNearbyDuplicate(nums, k));

    return 0;
}