diff --git a/notes/19. 正则表达式匹配.md b/notes/19. 正则表达式匹配.md
index a0ad0053..cb722e96 100644
--- a/notes/19. 正则表达式匹配.md	
+++ b/notes/19. 正则表达式匹配.md	
@@ -1,6 +1,6 @@
 # 19. 正则表达式匹配
 
-[牛客网](https://www.nowcoder.com/practice/28970c15befb4ff3a264189087b99ad4?tpId=13&tqId=11205&tab=answerKey&from=cyc_github)
+[Leetcode](https://leetcode.cn/problems/zheng-ze-biao-da-shi-pi-pei-lcof/)
 
 ## 题目描述
 
@@ -13,22 +13,22 @@
 应该注意到，'.' 是用来当做一个任意字符，而 '\*' 是用来重复前面的字符。这两个的作用不同，不能把 '.' 的作用和 '\*' 进行类比，从而把它当成重复前面字符一次。
 
 ```java
-public boolean match(String str, String pattern) {
+public boolean articleMatch(String s, String p) {
 
-    int m = str.length(), n = pattern.length();
+    int m = s.length(), n = p.length();
     boolean[][] dp = new boolean[m + 1][n + 1];
 
     dp[0][0] = true;
     for (int i = 1; i <= n; i++)
-        if (pattern.charAt(i - 1) == '*')
+        if (p.charAt(i - 1) == '*')
             dp[0][i] = dp[0][i - 2];
 
     for (int i = 1; i <= m; i++)
         for (int j = 1; j <= n; j++)
-            if (str.charAt(i - 1) == pattern.charAt(j - 1) || pattern.charAt(j - 1) == '.')
+            if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.')
                 dp[i][j] = dp[i - 1][j - 1];
-            else if (pattern.charAt(j - 1) == '*')
-                if (pattern.charAt(j - 2) == str.charAt(i - 1) || pattern.charAt(j - 2) == '.') {
+            else if (p.charAt(j - 1) == '*')
+                if (p.charAt(j - 2) == s.charAt(i - 1) || p.charAt(j - 2) == '.') {
                     dp[i][j] |= dp[i][j - 1]; // a* counts as single a
                     dp[i][j] |= dp[i - 1][j]; // a* counts as multiple a
                     dp[i][j] |= dp[i][j - 2]; // a* counts as empty