LeetCode:Factorial Trailing Zeroes

Posted 2020-07-26 walker lee

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.

Subscribe to see which companies asked this question

Hide Tags

 Math

Hide Similar Problems

 (H) Number of Digit One

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;
    }
};