[Swift]LeetCode786. 第 K 个最小的素数分数 | K-th Smallest Prime Fraction

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A sorted list A contains 1, plus some number of primes.  Then, for every p < q in the list, we consider the fraction p/q.

What is the K-th smallest fraction considered?  Return your answer as an array of ints, where answer[0] = p and answer[1] = q.

Examples:
Input: A = [1, 2, 3, 5], K = 3
Output: [2, 5]
Explanation:
The fractions to be considered in sorted order are:
1/5, 1/3, 2/5, 1/2, 3/5, 2/3.
The third fraction is 2/5.

Input: A = [1, 7], K = 1
Output: [1, 7]

Note:

• A will have length between 2 and 2000.
• Each A[i] will be between 1 and 30000.
• K will be between 1 and A.length * (A.length - 1) / 2.

---

一个已排序好的表 A，其包含 1 和其他一些素数.  当列表中的每一个 p<q 时，我们可以构造一个分数 p/q 。

那么第 k 个最小的分数是多少呢?  以整数数组的形式返回你的答案, 这里 answer[0] = p 且 answer[1] = q.

示例:
输入: A = [1, 2, 3, 5], K = 3
输出: [2, 5]
解释:
已构造好的分数,排序后如下所示:
1/5, 1/3, 2/5, 1/2, 3/5, 2/3.
很明显第三个最小的分数是 2/5.

输入: A = [1, 7], K = 1
输出: [1, 7]

注意:

• A 的取值范围在 2 — 2000.
• 每个 A[i] 的值在 1 —30000.
• K 取值范围为 1 —A.length * (A.length - 1) / 2

---

Runtime: 64 ms

Memory Usage: 19.1 MB

class Solution {
    func kthSmallestPrimeFraction(_ A: [Int], _ K: Int) -> [Int] {
        var left:Double = 0
        var right:Double = 1.0
        var p:Int = 0
        var q:Int = 1
        var cnt:Int = 0
        var n:Int = A.count
        while(true)
        {
            var mid:Double = left + (right - left) / 2.0
            cnt = 0
            p = 0
            var j:Int = 0
            for i in 0..<n
            {
                while(j < n && Double(A[i]) > mid * Double(A[j]))
                {
                    j += 1
                }
                cnt += n - j
                if j < n && p * A[j] < q * A[i]
                {
                    p = A[i]
                    q = A[j]
                }                
            }
            if cnt == K {return [p,q]}
            else if cnt < K {left = mid}
            else {right = mid}
        }
    }
}