作者：xuanchen | 来源：互联网 | 2023-10-10 12:31
FactorialTrailingZeroesGivenanintegern,returnthenumberoftrailingzeroesinn!.Note:Yoursoluti
Factorial Trailing Zeroes
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
对n!做质因数分解n!=2x*3y*5z*...
显然0的个数等于min(x,z)，并且min(x,z)==z
解法一：
从1到n中提取所有的5
class Solution {
public:
int trailingZeroes(int n) {
int ret = 0;
for(int i = 1; i <= n; i ++)
{
int tmp = i;
while(tmp%5 == 0)
{
ret ++;
tmp /= 5;
}
}
return ret;
}
};
解法二：
由上述分析可以看出，起作用的只有被5整除的那些数。能不能只对这些数进行计数呢？
存在这样的规律：[n/k]代表1~n中能被k整除的个数。
因此解法一可以转化为解法二
class Solution {
public:
int trailingZeroes(int n) {
int ret = 0;
while(n)
{
ret += n/5;
n /= 5;
}
return ret;
}
};
【LeetCode】Factorial Trailing Zeroes (2 solutions)
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