[Swift]LeetCode793. 阶乘函数后K个零 | 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的数量。（回想一下 x! = 1 * 2 * 3 * ... * x，且0! = 1）

例如， f(3) = 0 ，因为3! = 6的末尾没有0；而 f(11) = 2 ，因为11!= 39916800末端有2个0。给定 K，找出多少个非负整数x ，有 f(x) = K 的性质。

示例 1:
输入:K = 0
输出:5
解释: 0!, 1!, 2!, 3!, and 4! 均符合 K = 0 的条件。

示例 2:
输入:K = 5
输出:0
解释:没有匹配到这样的 x!，符合K = 5 的条件。

注意：

• K是范围在 [0, 10^9] 的整数。

---

Runtime: 4 ms

Memory Usage: 18.7 MB

class Solution {
    func preimageSizeFZF(_ K: Int) -> Int {
        if K < 5 {return 5}
        var base:Int = 1
        while(base * 5 + 1 <= K)
        {
            base = base * 5 + 1
        }
        if K / base == 5
        {
            return 0
        }
        return preimageSizeFZF(K % base)
    }
}

---

4ms

class Solution {    
    func preimageSizeFZF(_ K: Int) -> Int {
        var start = 1, end = Int.max, mid = start + (end - start) / 2
        
        while start < end {
            let candidate = trailingZeroes(mid)
            if candidate > K {
                end = mid - 1
            } else if candidate < K {
                start = mid + 1
            } else {
                return 5
            }
            mid = start + (end - start) / 2
        }
        return 0
    }
    
    func trailingZeroes(_ n: Int) -> Int {
        var n = n, current = 0
        while n > 1 {
            n /= 5
            current += n
        }
        
        return current
    }
}