diff --git a/src/main/java/leetcode/editor/cn/FactorialZerosLcci.java b/src/main/java/leetcode/editor/cn/FactorialZerosLcci.java
new file mode 100644
index 0000000..2625840
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+++ b/src/main/java/leetcode/editor/cn/FactorialZerosLcci.java
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+//设计一个算法，算出 n 阶乘有多少个尾随零。
+//
+// 示例 1:
+//
+// 输入: 3
+//输出: 0
+//解释: 3! = 6, 尾数中没有零。
+//
+// 示例 2:
+//
+// 输入: 5
+//输出: 1
+//解释: 5! = 120, 尾数中有 1 个零.
+//
+// 说明: 你算法的时间复杂度应为 O(log n) 。
+// Related Topics 数学 👍 66 👎 0
+
+package leetcode.editor.cn;
+
+//面试题 16.05:阶乘尾数
+public class FactorialZerosLcci {
+ public static void main(String[] args) {
+ Solution solution = new FactorialZerosLcci().new Solution();
+ // TO TEST
+ }
+
+ //leetcode submit region begin(Prohibit modification and deletion)
+ class Solution {
+ public int trailingZeroes(int n) {
+ int count = 0;
+ while (n >= 5) {
+ count += n / 5;
+ n /= 5;
+ }
+ return count;
+ }
+ }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
diff --git a/src/main/java/leetcode/editor/cn/doc/content/FactorialZerosLcci.md b/src/main/java/leetcode/editor/cn/doc/content/FactorialZerosLcci.md
new file mode 100644
index 0000000..a521799
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/content/FactorialZerosLcci.md
@@ -0,0 +1,16 @@
+设计一个算法，算出 n 阶乘有多少个尾随零。
+
+示例 1:
+
+输入: 3
+输出: 0
+解释: 3! = 6, 尾数中没有零。
+
+示例 2:
+
+输入: 5
+输出: 1
+解释: 5! = 120, 尾数中有 1 个零.
+
+说明: 你算法的时间复杂度应为 O(log n) 。
+
👍 66👎 0
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