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Update 19. 正则表达式匹配.md

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题目链接改为力扣中的对应题目，题目描述和代码按该题的接口调整。
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2026-10-28 13:45:03 +08:00
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# 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