Problem:
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.
思路：
Solution (C++):
int trailingZeroes(int n) {
if (n < 5) return 0;
return n/5 + trailingZeroes(n/5);
}
性能：
Runtime: 4 ms Memory Usage: 5.8 MB
思路：
Solution (C++):
int trailingZeroes(int n) {
int ans = 0;
for (long long i = 5; i <= n; i *= 5) {
ans += n / i;
}
return ans;
}
性能：
Runtime: 4 ms Memory Usage: 6 MB
思路：
Solution (C++):
int trailingZeroes(int n) {
int ans = 0;
while (n/5) {
ans += n/5;
n /= 5;
}
return ans;
}
性能：
Runtime: 0 ms Memory Usage: 5.9 MB
