Java computes the number of n factorial end 0-leetcode 172 factorial Trailing Zeroes

Topic

Given an integer n, return the number of trailing zeroes in N!.

Note:your solution should is in logarithmic time complexity.

Analysis

Note that the time complexity of the logarithm is solved, then it is impossible to roughly calculate the factorial of N and then look at the number of 0 at the end.

So careful analysis, N! = 1 * 2 * 3 * ... * N and the number of the end 0 is only related to the number of the multipliers in 5 and 2, because each occurrence of a pair of 5 and 2 will produce a 10 then N! There must be a 0 at the end. However, further analysis will find that 2 of the number of factors is definitely greater than the number of 5, so we just need to find n! The total number of factor 5 can be.

1. Given n then N/5 will get all the 5*1, 5*2, 5*3 ... the number of

2.N/25 will get the number of all 25*1, 25*2, 25*3 ....

3.N/125 will get all 125*1, 125*2, 125*3 .... Number of

........

There may be doubt that 25 of the two 5,125 in 3 5 will be less, but careful observation will find that in the 1th step has been the case of 25, the 2nd step is equivalent to the other 5 statistics in, and so on

Code

1      Public int trailingzeroes (int  n) {2         int rs = 0; 3          while (n! = 0) {4             rs + = (n/5); 5             n/= 5; 6         }7         return  rs; 8     }

And that's the result.

Java computes the number of n factorial end 0-leetcode 172 factorial Trailing Zeroes