leetcode 172. Factorial Trailing Zeroes

Given an integer n, return the number of trailing zeroes in n!.

Example 1:

Input: 3
Output: 0
Explanation: 3! = 6, no trailing zero.

Example 2:

Input: 5
Output: 1
Explanation: 5! = 120, one trailing zero.

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class Solution(object):     def trailingZeroes(self, n):         # 15!=3, 15/5=3         # 25!=6, 25/5=5+5^2|1         # count 5 number         if n < 5:             return 0         return n/5 + self.trailingZeroes(n/5)

 思考这类问题还是要从上而下思考，先用递归思路去解，最后修改循环或者dp。

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class Solution(object):     def trailingZeroes(self, n):         # 15!=3, 15/5=3         # 25!=6, 25/5=5+5^2|1         # count 5 number         ans = 0         while n >= 5:             ans += n/5             n = n/5         return ans

 另外的解法：

Example Three

 

By given number 4617.

 

5^1 : 4617 ÷ 5 = 923.4, so we get 923 factors of 5

 

5^2 : 4617 ÷ 25 = 184.68, so we get 184 additional factors of 5

 

5^3 : 4617 ÷ 125 = 36.936, so we get 36 additional factors of 5

 

5^4 : 4617 ÷ 625 = 7.3872, so we get 7 additional factors of 5

 

5^5 : 4617 ÷ 3125 = 1.47744, so we get 1 more factor of 5

 

5^6 : 4617 ÷ 15625 = 0.295488, which is less than 1, so stop here.

 

Then 4617! has 923 + 184 + 36 + 7 + 1 = 1151 trailing zeroes.

 

C/C++ code

 

int trailingZeroes(int n) {
    int result = 0;
    for(long long i=5; n/i>0; i*=5){
        result += (n/i);
    }
    return result;
}