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：
代码语言：javascript复制 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;
} 
