- leetcode:172. Factorial Trailing
- [String]172. Factorial Trailing
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- 172. Factorial Trailing Zeroes
- LeetCode 172. Factorial Trailing
- Leetcode 172. Factorial Trailing
题目：172. Factorial Trailing Zeroes
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
n！得到的数末尾有多少零。
2 ＊ 5 ＝ 10 带来一个0；所以只需计算n！里的25的pair。2很多（双数都可以）所以计算5得出pair数，即n／5。
然而考虑5的n次幂，以25为例，是55，也就是可以分别和两个2相乘得到0，所以实际要计算n/5/5...直到商为0，即不能被5整除。结果为n/5+n/5/5...之和
class Solution {
public int trailingZeroes(int n) {
int res = 0;
while (n/5 !=0){
res += n/5;
n /= 5;
}
return res;
}
}
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