Given an integer array, return the k-th smallest distance among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B.

Example 1:
Input:
nums = [1,3,1]
k = 1
Output: 0
Explanation:
Here are all the pairs:
(1,3) -> 2
(1,1) -> 0
(3,1) -> 2
Then the 1st smallest distance pair is (1,1), and its distance is 0.
Note:
2 <= len(nums) <= 10000.
0 <= nums[i] < 1000000.
1 <= k <= len(nums) * (len(nums) - 1) / 2.

本題題意很簡單，直接暴力去做即可

程式碼如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
 

#include <numeric>
#include <cmath>
#include <regex>

using namespace std;



class Solution 
{
public:
    int smallestDistancePair(vector<int>& nums, int k) 
    {
        int n = nums.size(), N = 1000000;
        vector<int> count(N, 0);
        for (int i = 0; i < n; i++)
        {
            for
 
 (int j = i + 1; j < n; j++)
            {
                count[abs(nums[i] - nums[j])]++;
            }
        }

        for (int i = 0; i < count.size(); i++)
        {
            if (count[i] >= k)
                return i;
            else
                k -= count[i];
        }
        return -1;
    }
};