[Swift]LeetCode719. 找出第 k 小的距离对 | Find K-th Smallest Pair Distance

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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:

1. 2 <= len(nums) <= 10000.
2. 0 <= nums[i] < 1000000.
3. 1 <= k <= len(nums) * (len(nums) - 1) / 2.

---

给定一个整数数组，返回所有数对之间的第 k 个最小距离。一对 (A, B) 的距离被定义为 A 和 B 之间的绝对差值。

示例 1:

输入：
nums = [1,3,1]
k = 1
输出：0 
解释：
所有数对如下：
(1,3) -> 2
(1,1) -> 0
(3,1) -> 2
因此第 1 个最小距离的数对是 (1,1)，它们之间的距离为 0。

提示:

1. 2 <= len(nums) <= 10000.
2. 0 <= nums[i] < 1000000.
3. 1 <= k <= len(nums) * (len(nums) - 1) / 2.

---

Runtime: 68 ms

Memory Usage: 19.6 MB

class Solution {
    func smallestDistancePair(_ nums: [Int], _ k: Int) -> Int {
        var nums = nums.sorted(by:<)
        var n:Int = nums.count
        var left:Int = 0
        var right:Int = nums.last! - nums[0]
        while (left < right) 
        {
            var mid:Int = left + (right - left) / 2
            var cnt:Int = 0
            var start:Int = 0
            for i in 0..<n
            {
                while (start < n && nums[i] - nums[start] > mid)
                {
                    start += 1
                }
                cnt += i - start
            }
            if cnt < k
            {
                left = mid + 1
            }
            else
            {
                right = mid
            }
        }
        return right       
    }
}

---

72ms

class Solution {
  func smallestDistancePair(_ nums: [Int], _ k: Int) -> Int {
    var sorted = nums
    sorted.sort { $0 < $1 }
    
    var lo = 0
    var hi = sorted[sorted.count - 1] - sorted[0]
    while lo < hi {
      let m = lo + ((hi - lo) / 2)
      var count = 0
      var j = 0
      for i in 0 ..< sorted.count - 1 {
        while j < sorted.count && sorted[j] - sorted[i] <= m {
          j += 1
        }
        count += j - i - 1
      }
      if count < k {
        lo = m + 1
      } else {
        hi = m
      }
    }
    return lo
  }
}

---

88ms

class Solution {
    func smallestDistancePair(_ nums: [Int], _ k: Int) -> Int {
        let nums = nums.sorted()
        var low = 0, high = abs((nums.first ?? 0) - (nums.last ?? 0))
        while low < high {
            let m = (low + high) / 2
            var count = 0
            var j = 1
            for i in 0..<nums.count-1 {
                while j < nums.count && nums[j] - nums[i] <= m { j += 1 }
                count += j-i-1;
            }
            if count < k {
                low = m + 1
            } else {
                high = m
            }
        }
        return low
    }
}