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
.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;
}
};