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 = 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.

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。

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;
}
};