diff --git a/notes/31. 栈的压入、弹出序列.md b/notes/31. 栈的压入、弹出序列.md
index 69817d96..400a8448 100644
--- a/notes/31. 栈的压入、弹出序列.md	
+++ b/notes/31. 栈的压入、弹出序列.md	
@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[牛客网](https://www.nowcoder.com/practice/d77d11405cc7470d82554cb392585106?tpId=13&tqId=11174&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[Leetcode](https://leetcode.cn/problems/zhan-de-ya-ru-dan-chu-xu-lie-lcof/)
 
 ## 题目描述
 
@@ -15,13 +15,12 @@
 使用一个栈来模拟压入弹出操作。每次入栈一个元素后，都要判断一下栈顶元素是不是当前出栈序列 popSequence 的第一个元素，如果是的话则执行出栈操作并将 popSequence 往后移一位，继续进行判断。
 
 ```java
-public boolean IsPopOrder(int[] pushSequence, int[] popSequence) {
-    int n = pushSequence.length;
+public boolean validateBookSequences(int[] putIn, int[] takeOut) {
+    int n = putIn.length;
     Stack<Integer> stack = new Stack<>();
     for (int pushIndex = 0, popIndex = 0; pushIndex < n; pushIndex++) {
-        stack.push(pushSequence[pushIndex]);
-        while (popIndex < n && !stack.isEmpty() 
-                && stack.peek() == popSequence[popIndex]) {
+        stack.push(putIn[pushIndex]);
+        while (popIndex < n && !stack.isEmpty() && stack.peek() == takeOut[popIndex]) {
             stack.pop();
             popIndex++;
         }