172. Factorial Trailing Zeroes -- 求n的阶乘末尾有几个0

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

Note: Your solution should be in logarithmic time complexity.

（1）

class Solution {
public:
int trailingZeroes(int n) {
int ans = 0;
for(long long i = 5; n / i; i *= 5)
{
ans += n / i;
}
return ans;
}
};

（2）

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