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 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

.
题意：给出一个数组，定义任意两个的距离为它们差的绝对值。求第k大的距离。代码：class Solution {
public:
int smallestDistancePair(vector<int>& nums, int k) {
sort(nums.begin(),nums.end());
int s=0,t=nums[nums.size()-1]-nums[0];
while(s<t)
{
int mid=(s+t)>>1;
int left=0,count=0;
for(int right=0;right<nums.size();right++)
{
while(nums[right]-nums[left]>mid)
left++;
count+=right-left;
}
if(count<k)
s=mid+1;
else
t=mid;
}
return s;
}
};