LeetCode:Factorial Trailing Zeroes
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.
借鉴别人的 http://www.cnblogs.com/ganganloveu/p/4193373.html 代码简洁高效
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
证明:
对于阶乘而言,也就是1*2*3*...*n
[n/k]代表1~n中能被k整除的个数
那么很显然
[n/2] > [n/5] (左边是逢2增1,右边是逢5增1)
[n/2^2] > [n/5^2](左边是逢4增1,右边是逢25增1)
……
[n/2^p] > [n/5^p](左边是逢2^p增1,右边是逢5^p增1)
随着幂次p的上升,出现2^p的概率会远大于出现5^p的概率。
因此左边的加和一定大于右边的加和,也就是n!质因数分解中,2的次幂一定大于5的次幂
1 class Solution { 2 public: 3 int trailingZeroes(int n) { 4 int result=0; 5 while(n) 6 { 7 result+=n/5; 8 n/=5; 9 } 10 return result; 11 12 } 13 };