leetcode 719. Find K-th Smallest Pair Distance c++

原创于 2019-04-11 20:25:57 发布 · 181 阅读

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本文介绍了一种解决寻找整数数组中第k小距离对问题的高效算法。通过排序和二分查找，该算法能在O(n log n)的时间复杂度内找到所有数对中距离最小的前k个。示例中，对于输入数组[1,3,1]和k=1，输出为0，即最近的数对(1,1)的距离。

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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.

class Solution {
public:
    int smallestDistancePair(vector<int>& nums, int k) 
    {
        int len = nums.size();
        sort(nums.begin(),nums.end());
        int left = 0;
        int right = nums[len-1]-nums[0];
        int mid=0;
        while(left<right)
        {
            mid = left + (right-left)/2;
            int lowl=0;
            int count = 0;
            for(int upl=0;upl<len;upl++)
            {
                while(nums[upl]-nums[lowl]>mid) lowl++;
                count += upl - lowl;
            }
            if(count<k) left = mid+1;
            else right = mid;
        }
        return left;
    }
};