leetcode 172 Factorial Trailing Zeroes

原创于 2016-06-19 14:16:37 发布 · 239 阅读

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本文介绍了一种计算任意整数n的阶乘结果中尾随0的数量的方法，并且强调了该算法的时间复杂度为对数级别。通过迭代地除以5的倍数来计算尾随0的数量。

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.

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class Solution {
public:
	int trailingZeroes(int n) {
		if(n < 1) return 0;
		int cnt = 0;

		while(n >= 5) {
			cnt += n/5;
			n = n/5;
		}

		return cnt;
	}
}