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;    }};