[Swift]LeetCode668.乘法表中第k小的数|KthSmallestNumberinMultiplicationTable ...

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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:

The m and n will be in the range [1, 30000].
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：

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

Runtime: 52 ms

Memory Usage: 18.5 MB

 1 class Solution {
 2     func findKthNumber(_ m: Int, _ n: Int, _ k: Int) -> Int {
 3         var left:Int = 1
 4         var right:Int = m * n
 5         while (left < right)
 6         {
 7             var mid:Int = left + (right - left) / 2
 8             var cnt:Int = 0
 9             var i:Int = m
10             var j:Int = 1
11             while (i >= 1 && j <= n)
12             {
13                 var t:Int = j
14                 j = (mid > n * i) ? n + 1 : (mid / i + 1)
15                 cnt += (j - t) * i
16                 i = mid / j
17             }
18             if cnt < k
19             {
20                 left = mid + 1
21             }
22             else
23             {
24                 right = mid
25             }                    
26         }
27         return right
28     }
29 }

80ms

 1 class Solution {
 2     func findKthNumber(_ m: Int, _ n: Int, _ k: Int) -> Int {
 3         var low = 1, high = m * n
 4         while (low <= high) {
 5             let mid = low + (high - low) / 2
 6             let count = helper(m, n, mid)
 7             if count >= k { high = mid - 1 }
 8             else { low = mid + 1 }
 9         }
10         return low
11     }
12     
13     func helper(_ m: Int, _ n: Int, _ num: Int) -> Int {
14         var count = 0
15         let mi = min(m, n), ma = max(m, n)
16         for i in 1...mi {
17             count += min(num / i, ma)
18         }
19         return count
20     }
21 }

92ms

 1 class Solution {
 2     func findKthNumber(_ m: Int, _ n: Int, _ k: Int) -> Int {
 3         var low = 1
 4         var high = k
 5         
 6         while low < high {
 7             let mid = (low + high) / 2
 8             var count = 0
 9             for i in 1...m {
10                 count += min(mid/i, n)
11             }
12             
13             if count < k {
14                 low = mid + 1
15             } else {
16                 high = mid
17             }
18         }
19         
20         return low
21     }
22 }

18348kb

 1 class Solution {
 2     func findKthNumber(_ m: Int, _ n: Int, _ k: Int) -> Int {
 3         var low = 1
 4         var high = k
 5         
 6         while low < high {
 7             let mid = (low + high) / 2
 8             var count = 0
 9             for i in 1...m {
10                 count += min(mid/i, n)
11             }
12             
13             if count >= k {
14                 high = mid
15             } else {
16                 low = mid + 1
17             }
18         }
19         
20         return high
21     }
22 }