719. Find K-th Smallest Pair Distance

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:

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.

cpp_code:

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class Solution {
public:
    int f_abs(int x){
        return x > 0 ? x : -x;
    }
    int count(vector<int>& nums, int mid){
        int cnt = 0;
        int n = nums.size();
        int j = 0;
        //寻找i前面与nums[i]差值刚好不大于mid的j，找i+1的时候直接就在j的基础上往后找，利用了滑动窗口的思想
        for(int i = 0; i < n; i++){
            while(j < i && (nums[i] - nums[j]) > mid) j++;
            cnt += (i - j);
        }
        return cnt;
    }
    int smallestDistancePair(vector<int>& nums, int k) {
        int n = nums.size();
        sort(nums.begin(),nums.end());
        int left = 0;
        int right = nums[n-1] - nums[0];
        //二分，若小于差值mid的个数小于k，则需要将差值增大，反之，则减小。
        while(left < right){
            int mid = (right + left) / 2;
            if(count(nums, mid) < k){
                left = mid + 1;
            }else{
                right = mid;
            }
        }
        return left;
    }
};