标签：res python bsp ret 大于 inpu out else amp
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.
给一个整数n，返回n的阶乘末尾0的个数。
找乘数中10的个数，而10可分解为2和5，而2的数量远大于5的数量，所以找出5的个数。
解法1：迭代Iterative
解法2: 递归Recursive
Java:
public class Solution {
public int trailingZeroes(int n) {
int res = 0;
while (n > 0) {
res += n / 5;
n /= 5;
}
return res;
}
}
Java:
public class Solution {
public int trailingZeroes(int n) {
return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
}
}
Python:
class Solution:
# @return an integer
def trailingZeroes(self, n):
result = 0
while n > 0:
result += n / 5
n /= 5
return result
Python:
class Solution(object):
def trailingZeroes(self, n):
"""
:type n: int
:rtype: int
"""
return 0 if n == 0 else n / 5 + self.trailingZeroes(n / 5)
C++:
class Solution {
public:
int trailingZeroes(int n) {
int res = 0;
while (n) {
res += n / 5;
n /= 5;
}
return res;
}
};
C++:
class Solution {
public:
int trailingZeroes(int n) {
return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
}
};
[LeetCode] 172. Factorial Trailing Zeroes 求阶乘末尾零的个数
标签：res python bsp ret 大于 inpu out else amp
原文地址：https://www.cnblogs.com/lightwindy/p/9736240.html
