[Swift]LeetCode668. 乘法表中第k小的数 | Kth Smallest Number in Multiplication Table

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10492422.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Nearly every one have used the Multiplication Table. But could you find out the k-th smallest number quickly from the multiplication table?

Given the height m and the length n of a m * nMultiplication Table, and a positive integer k, you need to return the k-th smallest number in this table.

Example 1:

Input: m = 3, n = 3, k = 5
Output: 
Explanation: 
The Multiplication Table:
1	2	3
2	4	6
3	6	9

The 5-th smallest number is 3 (1, 2, 2, 3, 3). 

Example 2:

Input: m = 2, n = 3, k = 6
Output: 
Explanation: 
The Multiplication Table:
1	2	3
2	4	6

The 6-th smallest number is 6 (1, 2, 2, 3, 4, 6). 

Note:

1. The m and n will be in the range [1, 30000].
2. The k will be in the range [1, m * n]

---

几乎每一个人都用 乘法表。但是你能在乘法表中快速找到第k小的数字吗？

给定高度m 、宽度n 的一张 m * n的乘法表，以及正整数k，你需要返回表中第k 小的数字。

例 1：

输入: m = 3, n = 3, k = 5
输出: 3
解释: 
乘法表:
1	2	3
2	4	6
3	6	9

第5小的数字是 3 (1, 2, 2, 3, 3).

例 2：

输入: m = 2, n = 3, k = 6
输出: 6
解释: 
乘法表:
1	2	3
2	4	6

第6小的数字是 6 (1, 2, 2, 3, 4, 6).

注意：

1. m 和 n 的范围在 [1, 30000] 之间。
2. k 的范围在 [1, m * n] 之间。

---

Runtime: 52 ms

Memory Usage: 18.5 MB

class Solution {
    func findKthNumber(_ m: Int, _ n: Int, _ k: Int) -> Int {
        var left:Int = 1
        var right:Int = m * n
        while (left < right)
        {
            var mid:Int = left + (right - left) / 2
            var cnt:Int = 0
            var i:Int = m
            var j:Int = 1
            while (i >= 1 && j <= n)
            {
                var t:Int = j
                j = (mid > n * i) ? n + 1 : (mid / i + 1)
                cnt += (j - t) * i
                i = mid / j
            }
            if cnt < k
            {
                left = mid + 1
            }
            else
            {
                right = mid
            }                    
        }
        return right
    }
}

---

80ms

class Solution {
    func findKthNumber(_ m: Int, _ n: Int, _ k: Int) -> Int {
        var low = 1, high = m * n
        while (low <= high) {
            let mid = low + (high - low) / 2
            let count = helper(m, n, mid)
            if count >= k { high = mid - 1 }
            else { low = mid + 1 }
        }
        return low
    }
    
    func helper(_ m: Int, _ n: Int, _ num: Int) -> Int {
        var count = 0
        let mi = min(m, n), ma = max(m, n)
        for i in 1...mi {
            count += min(num / i, ma)
        }
        return count
    }
}

---

92ms

class Solution {
    func findKthNumber(_ m: Int, _ n: Int, _ k: Int) -> Int {
        var low = 1
        var high = k
        
        while low < high {
            let mid = (low + high) / 2
            var count = 0
            for i in 1...m {
                count += min(mid/i, n)
            }
            
            if count < k {
                low = mid + 1
            } else {
                high = mid
            }
        }
        
        return low
    }
}

---

18348kb

class Solution {
    func findKthNumber(_ m: Int, _ n: Int, _ k: Int) -> Int {
        var low = 1
        var high = k
        
        while low < high {
            let mid = (low + high) / 2
            var count = 0
            for i in 1...m {
                count += min(mid/i, n)
            }
            
            if count >= k {
                high = mid
            } else {
                low = mid + 1
            }
        }
        
        return high
    }
}