LWC 56：719. Find K-th Smallest Pair Distance

LWC 56：719. Find K-th Smallest Pair Distance

传送门：719. Find K-th Smallest Pair Distance

Problem:

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.

思路：

排序+二分+尺取法，实际上对数组排序后，小于某个数的对数可以在线性时间内求出（尺取法的思想），那么意味着问题需要猜值验证，所以二分，代码如下：

public int smallestDistancePair(int[] nums, int k) {
int n = nums.length;
Arrays.sort(nums);
int lf = -1;
int rt = nums[n - 1] - nums[0];

while (rt - lf > 1) {
int mid = lf + (rt - lf) / 2;
if (count(nums, mid) < k) {
lf = mid;
}
else {
rt = mid;
}
}
return rt;
}

public int count(int[] nums, int mid) {  // <= mid 的个数
int n = nums.length;
int cnt = 0;

int j = 0;

for (int i = 1; i < n; ++i) {
while (j < i && nums[i] - nums[j] > mid) j++;
cnt += i - j;
}

return cnt;
}