diff --git a/src/main/java/leetcode/editor/cn/FeiBoNaQiShuLieLcof.java b/src/main/java/leetcode/editor/cn/FeiBoNaQiShuLieLcof.java
new file mode 100644
index 0000000..a61364f
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/FeiBoNaQiShuLieLcof.java
@@ -0,0 +1,65 @@
+//写一个函数，输入 n ，求斐波那契（Fibonacci）数列的第 n 项（即 F(N)）。斐波那契数列的定义如下：
+//
+//
+//F(0) = 0, F(1) = 1
+//F(N) = F(N - 1) + F(N - 2), 其中 N > 1.
+//
+// 斐波那契数列由 0 和 1 开始，之后的斐波那契数就是由之前的两数相加而得出。
+//
+// 答案需要取模 1e9+7（1000000007），如计算初始结果为：1000000008，请返回 1。
+//
+//
+//
+// 示例 1：
+//
+//
+//输入：n = 2
+//输出：1
+//
+//
+// 示例 2：
+//
+//
+//输入：n = 5
+//输出：5
+//
+//
+//
+//
+// 提示：
+//
+//
+// 0 <= n <= 100
+//
+// Related Topics 记忆化搜索 数学 动态规划 👍 223 👎 0
+
+package leetcode.editor.cn;
+
+//剑指 Offer 10- I:斐波那契数列
+class FeiBoNaQiShuLieLcof {
+ public static void main(String[] args) {
+ //测试代码
+ Solution solution = new FeiBoNaQiShuLieLcof().new Solution();
+ }
+
+ //力扣代码
+ //leetcode submit region begin(Prohibit modification and deletion)
+ class Solution {
+ public int fib(int n) {
+ if (n < 2) {
+ return n;
+ }
+ int f0 = 0;
+ int f1 = 1;
+ long result = 0;
+ for (int i = 2; i <= n; ++i) {
+ result = (f0 + f1) % 1000000007;
+ f0 = f1;
+ f1 = (int) result;
+ }
+ return (int) result;
+ }
+ }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
\ No newline at end of file
diff --git a/src/main/java/leetcode/editor/cn/doc/all.json b/src/main/java/leetcode/editor/cn/doc/all.json
index c68613d..f7772e3 100644
--- a/src/main/java/leetcode/editor/cn/doc/all.json
+++ b/src/main/java/leetcode/editor/cn/doc/all.json
@@ -1 +1 @@
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\ No newline at end of file
diff --git a/src/main/java/leetcode/editor/cn/doc/content/FeiBoNaQiShuLieLcof.md b/src/main/java/leetcode/editor/cn/doc/content/FeiBoNaQiShuLieLcof.md
new file mode 100644
index 0000000..4a2e65c
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/content/FeiBoNaQiShuLieLcof.md
@@ -0,0 +1,34 @@
+写一个函数，输入 n ，求斐波那契（Fibonacci）数列的第 n 项（即 F(N)）。斐波那契数列的定义如下：
+
+
+F(0) = 0, F(1) = 1
+F(N) = F(N - 1) + F(N - 2), 其中 N > 1.
+
+斐波那契数列由 0 和 1 开始，之后的斐波那契数就是由之前的两数相加而得出。
+
+答案需要取模 1e9+7（1000000007），如计算初始结果为：1000000008，请返回 1。
+
+
+
+示例 1：
+
+
+输入：n = 2
+输出：1
+
+
+示例 2：
+
+
+输入：n = 5
+输出：5
+
+
+
+
+提示：
+
+
+Related Topics
记忆化搜索数学动态规划
👍 223👎 0
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