LeetCode_172. Factorial Trailing Zeroes

172. Factorial Trailing Zeroes

Easy

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

Example 1:

Input: 3
Output: 0
Explanation: 3! = 6, no trailing zero.

Example 2:

Input: 5
Output: 1
Explanation: 5! = 120, one trailing zero.

Note: Your solution should be in logarithmic time complexity.

package leetcode.easy;

public class FactorialTrailingZeroes {
	@org.junit.Test
	public void test() {
		System.out.println(trailingZeroes(3));
		System.out.println(trailingZeroes(5));
	}

	public int trailingZeroes(int n) {
		int count = 0;
		while (n > 0) {
			n /= 5;
			count += n;
		}
		return count;
	}
}
原文地址:https://www.cnblogs.com/denggelin/p/11692012.html