题目描述
Given an integer n, return the number of trailing zeroes in n!.
Example1:
Input: 3 Output: 0 Explanation: 3! = 6, no trailing zero.
Example2:
Input: 5 Output: 1 Explanation: 5! = 120, one trailing zero.
NOTE:
- Your solution should be in logarithmic time complexity.
解题思路
- 求阶乘末尾0的个数
- 实质是求在乘的过程中，有多少个10
- 10可分解成5*2，其中2的个数又远大于5，即本题计算给定的数字含有5的个数
class Solution {
public:
int trailingZeroes(int n) {
int res = 0;
while(n){
res += n / 5;
n /= 5;
}
return res;
}
};
