LeetCode:Factorial Trailing Zeroes

Factorial Trailing Zeroes




Total Accepted: 61757  Total Submissions: 185808  Difficulty: Easy

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

Note: Your solution should be in logarithmic time complexity.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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思路:

1.题意求n!中后缀0的个数。

2.n!=1*2*3*...*n,中的0由(2^i) * (5^j)得来。

3.即要计算min(i,j)。

4.j<=i,直观的来看:i是逢2进1,j是逢5进1。固只需计算5的个数。

例如:10!=1*2*3*4*5*6*7*8*9*10 = ..*(2^4)*...*(5^2)..

求10!中5的个数,即求 k = n/5 + n/25 + n/125 + ....+n/5^j,其中j<=n。


c++ code:

class Solution {
public:
    int trailingZeroes(int n) {
        int x = 5;
        int ret = 0;
        while(x <= n) {
            ret += n/x;
            x *= 5;
        }
	 return ret;
    }
};


或:

class Solution {
public:
    int trailingZeroes(int n) {
        int ret = 0;
        while(n) {
            ret += n/5;
            n /= 5;
        }
	 return ret;
    }
};


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