算法第15周Find K-th Smallest Pair Distance[hard]

Description

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 = 1Output: 0 Explanation:Here are all the pairs:(1,3) -> 2(1,1) -> 0(3,1) -> 2Then 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.

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Analysis

这道题的意思就是找出第k小的距离，这个距离是数组任意两个数的差的绝对值。如果我们计算出这所有的差值，然后进行排序，找到第k小的距离，这种算法复杂度太高。我们可以采用二分法。
首先我们对数组进行排序，最大的数减最小的数即为距离的最大值，我们假设距离的最小值为0；
我们进行排序必须清楚：(nums为排序后数组）
nums[i]-nums[i-1] < nums[i+1]-nums[i-1];
剩下的就与二分法近似。

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Solution

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