LeetCode 172: Factorial Trailing Zeroes

LeetCode 172: Factorial Trailing Zeroes

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

 

// Question Description: given an integer n, return n! (Factorial of n) the number of suffixes 0. // Method 1: first obtain the factorial of n and then calculate the number of 0 at the end. This method is n when n is large! Class Solution {public: int trailingZeroes (int n) {if (n = 0) return 0; int m = 0; int cnt = 0; int k = factorial (n); while (k) {m = k % 10; k = k/10; if (m = 0) cnt ++; elsebreak ;} return cnt;} // calculate the factorial int factorial (int n1) of n {if (n1 = 0) return 1; else {return n1 * factorial (n1-1 );}}};

// Method 2: Consider n! Prime factor. The suffix 0 is always obtained by multiplying prime factor 2 and prime factor 5. If we can count the numbers of 2 and 5, the problem is solved. // Example: WHEN n = 5, 5! (2*2*2*3*5) contains one five and three two. Therefore, the number of suffixes 0 is 1. // N = 11, 11! (2 ^ 8) * (3 ^ 4) * (5 ^ 2) * 7) contains two 5 and eight 2. Therefore, the number of suffixes 0 is 2. // We can easily see that the number of 2 in the prime factor is always greater than or equal to 5, so we only need to count the number of 5. // How to calculate n! What is the number of all five in the prime factor? A simple method is to calculate floor (n/5 ). For example, 7! There is a 5, 10! There are two 5. // There is one more thing to consider. There are more than five numbers, such as 25,125. // For example, n = 25, n! = 25*24*23 *... * 15... * 10... * 5... * 1 = (5*5) * 24*23 *... * (5*3 )*... (5*2 )*... (5*1 )*... * 1, 25 of which can be regarded as 5*5, with a 5 more, it is very easy to add // to solve this problem. First, remove all single 5 for n limit 5, then 25, remove the additional 5, and so on. The formula for calculating the suffix 0 is as follows. // N! Number of suffixes 0 = n! Number of five in the prime factor = floor (n/5) + floor (n/25) + floor (n/125) + .... class Solution2 {public: int trailingZeroes (int n) {int res = 0; while (n) {res = res + n/5; n = n/5 ;} return res ;}};