[Swift]LeetCode793. 階乘函數後K個零 | Preimage Size of Factorial Zeroes Function

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Let f(x) be the number of zeroes at the end of x!. (Recall that x! = 1 * 2 * 3 * ... * x, and by convention, 0! = 1.)

For example, f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has 2 zeroes at the end. Given K, find how many non-negative integers xhave the property that f(x) = K.

Example 1:
Input: K = 0
Output: 5
Explanation: 0!, 1!, 2!, 3!, and 4! end with K = 0 zeroes.

Example 2:
Input: K = 5
Output: 0
Explanation: There is no x such that x! ends in K = 5 zeroes.

Note:

• K will be an integer in the range [0, 10^9].

---

 f(x) 是 x! 末尾是0的數量。（回想一下 x! = 1 * 2 * 3 * ... * x，且0! = 1）

例如， f(3) = 0 ，由於3! = 6的末尾沒有0；而 f(11) = 2 ，由於11!= 39916800末端有2個0。給定 K，找出多少個非負整數x ，有 f(x) = K 的性質。

示例 1:
輸入:K = 0
輸出:5
解釋: 0!, 1!, 2!, 3!, and 4! 均符合 K = 0 的條件。

示例 2:
輸入:K = 5
輸出:0
解釋:沒有匹配到這樣的 x!，符合K = 5 的條件。

注意：

• K是範圍在 [0, 10^9] 的整數。

---

Runtime: 4 ms

Memory Usage: 18.7 MB

 1 class Solution {
 2     func preimageSizeFZF(_ K: Int) -> Int {
 3         if K < 5 {return 5}
 4         var base:Int = 1
 5         while(base * 5 + 1 <= K)
 6         {
 7             base = base * 5 + 1
 8         }
 9         if K / base == 5
10         {
11             return 0
12         }
13         return preimageSizeFZF(K % base)
14     }
15 }

---

4ms

 1 class Solution {    
 2     func preimageSizeFZF(_ K: Int) -> Int {
 3         var start = 1, end = Int.max, mid = start + (end - start) / 2
 4         
 5         while start < end {
 6             let candidate = trailingZeroes(mid)
 7             if candidate > K {
 8                 end = mid - 1
 9             } else if candidate < K {
10                 start = mid + 1
11             } else {
12                 return 5
13             }
14             mid = start + (end - start) / 2
15         }
16         return 0
17     }
18     
19     func trailingZeroes(_ n: Int) -> Int {
20         var n = n, current = 0
21         while n > 1 {
22             n /= 5
23             current += n
24         }
25         
26         return current
27     }
28 }