LeetCode-219. Duplicate Element II present

Give you an array of integers  nums And an integer  k , Determine whether there are two... In the array Different indexes  i  and  j , Satisfy nums[i] == nums[j] And abs(i - j) <= k . If there is , return true ; otherwise , return false .

Example  1：

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

Input ：nums = [1,0,1,1], k = 1
Output ：true
Example 3：

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

Tips ：

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

#include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;

class Solution {
public:
	bool containsNearbyDuplicate(vector<int>& nums, int k) {
		unordered_set<int> mp;
		unordered_set<int>::iterator first;

		/*  Input verification  */
		if (nums.empty() || nums.size() == 1) {
			return false;
		}

		/*  A special case ,k Greater than the length of the array , Just judge once  */
		if (k >= nums.size()) {
			for (int i = 0; i < nums.size(); i++) {
				if (mp.count(nums[i]) == 0) {
					mp.insert(nums[i]);
				}
				else {
					return true;
				}
			}
			return false;
		}

		/* k Less than array length  */
		else {
			for (int i = 0; i <= k; i++) {
				if (mp.count(nums[i]) == 0) {
					mp.insert(nums[i]);
				}
				else {
					return true;
				}
			}
		}

		/*  Building sliding windows  */
		for (int i = k + 1; i < nums.size(); i++) {
			first = mp.find(nums[i - k - 1]);
			mp.erase(first);
			if (mp.count(nums[i]) == 0) {
				mp.insert(nums[i]);
			}
			else {
				return true;
			}
		}
		return false;
	}
};

int main() {
	vector<int> nums = { 1, 2, 3,1,2, 3 };
	Solution* ps = new Solution();
	cout << ps->containsNearbyDuplicate(nums, 2) << endl;
	system("pause");
	return 0;
}