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

Note: Your solution should be in logarithmic time complexity.

 

int trailingZeroes(int n) {
    if(n < 5)
        return 0;
    int k = 0;
    while(n)
    {
        n = n / 5;
        k += n;
    }
    return k;
}

转：http://www.jianshu.com/p/211618afc695