leetcode 719. Find K-th Smallest Pair Distance

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:

1. 2 <= len(nums) <= 10000.
2. 0 <= nums[i] < 1000000.
3. 1 <= k <= len(nums) * (len(nums) - 1) / 2.

1、二分答案。缩小范围。

2、滑动窗口可以减少运算量。

//要求返回第k小的差值，方法是 二分答案
//排序复杂度O(nlogn)
//二分+count的复杂度O(logm * m)
class Solution {
public:
    int smallestDistancePair(vector& nums, int k)
    {
        sort(nums.begin(), nums.end());
        int start = 0;
        int end = nums[nums.size() - 1] - nums[0];
        while (start + 1 < end)
        {
            int mid = (end - start) / 2 + start;
            if (count(nums, mid) >= k) //求的是 nums 之间的差值 小于等于 mid 的个数。
                end = mid;
            else
                start = mid;
        }
        //double check
        if ( count(nums, start) >= k)
            return start;
        else
            return end;
    }
    
    int count(vector& nums, int target) //差值 小于等于 target 的个数
    {
        //search
        //使用窗口思想，判断差值<=k的个数，r-l即表示[l,r]间间隔 target)
            {
                left++; //移动左边界
            }
            ret += right - left;
        }
        return ret;
    }
};