Factorial Trailing Zeroes(leetcode172)

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.

Note: Your solution should be in logarithmic time complexity.

開始寫了一個方法，結果超時了，以下：

public static int trailingZeroes(int n) {

    int num = 0;
    int num5 = 0;
    for (int i=1;i<=n;i++){
        int nu = i;
        while(nu%5==0 && nu >0){
            num5++;
            nu = nu/5;
        }
    }
    return num5;
}

改爲如下的方法能夠了：

public static int trailingZeroes(int n) {
    int count = 0;
    while(n > 1){
        n /= 5;
        count += n;
    }
    return count;
}

題目地址：https://leetcode.com/problems/factorial-trailing-zeroes/