diff --git a/notes/16. 数值的整数次方.md b/notes/16. 数值的整数次方.md
index b20367e2..b617a4f8 100644
--- a/notes/16. 数值的整数次方.md	
+++ b/notes/16. 数值的整数次方.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[牛客网](https://www.nowcoder.com/practice/1a834e5e3e1a4b7ba251417554e07c00?tpId=13&tqId=11165&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[Leetcode](https://leetcode.cn/problems/shu-zhi-de-zheng-shu-ci-fang-lcof/)
 
 ## 题目描述
 
@@ -24,19 +24,17 @@
 因为 (x\*x)<sup>n/2</sup> 可以通过递归求解，并且每次递归 n 都减小一半，因此整个算法的时间复杂度为 O(logN)。
 
 ```java
-public double Power(double x, int n) {
-    boolean isNegative = false;
-    if (n < 0) {
-        n = -n;
-        isNegative = true;
+public double myPow(double x, int n) {
+    long b = n;
+    if (b < 0) {
+        x = 1 / x;
+        b = -b;
     }
-    double res = pow(x, n);
-    return isNegative ? 1 / res : res;
+    return pow(x, b);
 }
 
-private double pow(double x, int n) {
+private double pow(double x, long n) {
     if (n == 0) return 1;
-    if (n == 1) return x;
     double res = pow(x, n / 2);
     res = res * res;
     if (n % 2 != 0) res *= x;