[LeetCode#172]Factorial Trailing Zeroes

Problem:

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.

Analysis:

The mathematical problem is always so tricky and easy, as long as you master the inherent theory behind it.
Reference:
https://en.wikipedia.org/wiki/Trailing_zero
Theory:
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, is simply the multiplicity of the prime factor 5 in n!.
http://bookshadow.com/weblog/2014/12/30/leetcode-factorial-trailing-zeroes/

Reason: 
The trailing zeros were actually caused by the 10 (which the the muliplicity of prime 5 and prime 2). And 
n = 5: There is one 5 and 3 2s in prime factors of 5! (2 * 2 * 2 * 3 * 5). So count of trailing 0s is 1.
n = 11: There are two 5s and three 2s in prime factors of 11! (2 8 * 34 * 52 * 7). So count of trailing 0s is 2.

The number of 2s in prime factors is apparenlty equal or larger than that of 5s in prime factors. Thus we only need to count the number of 5s.
Note: how to covert the equation(algorithm into program)!


The magic characteristics behind prime factors:
When we represent an integer into the prime factors form. 
1. The form must be unique. (since no prime factor is divideable)
2. We can gain some useful information from the prime factors. (like the number of trailing zero).

Solution:

public class Solution {
    public int trailingZeroes(int n) {
        int total = 0;
        while ( n / 5 != 0) {
            total += n / 5;
            n = n / 5;
        }
        return total;
    }
}

 

转载于:https://www.cnblogs.com/airwindow/p/4774319.html