48 lines
1.9 KiB
Java
48 lines
1.9 KiB
Java
# 19. 正则表达式匹配
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[NowCoder](https://www.nowcoder.com/practice/45327ae22b7b413ea21df13ee7d6429c?tpId=13&tqId=11205&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
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## 题目描述
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请实现一个函数用来匹配包括 '.' 和 '\*' 的正则表达式。模式中的字符 '.' 表示任意一个字符，而 '\*' 表示它前面的字符可以出现任意次（包含 0 次）。
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在本题中，匹配是指字符串的所有字符匹配整个模式。例如，字符串 "aaa" 与模式 "a.a" 和 "ab\*ac\*a" 匹配，但是与 "aa.a" 和 "ab\*a" 均不匹配。
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## 解题思路
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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。
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```java
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public boolean match(char[] str, char[] pattern) {
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int m = str.length, n = pattern.length;
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boolean[][] dp = new boolean[m + 1][n + 1];
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dp[0][0] = true;
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for (int i = 1; i <= n; i++)
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if (pattern[i - 1] == '*')
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dp[0][i] = dp[0][i - 2];
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for (int i = 1; i <= m; i++)
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for (int j = 1; j <= n; j++)
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if (str[i - 1] == pattern[j - 1] || pattern[j - 1] == '.')
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dp[i][j] = dp[i - 1][j - 1];
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else if (pattern[j - 1] == '*')
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if (pattern[j - 2] == str[i - 1] || pattern[j - 2] == '.') {
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dp[i][j] |= dp[i][j - 1]; // a* counts as single a
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dp[i][j] |= dp[i - 1][j]; // a* counts as multiple a
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dp[i][j] |= dp[i][j - 2]; // a* counts as empty
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} else
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dp[i][j] = dp[i][j - 2]; // a* only counts as empty
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return dp[m][n];
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}
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```
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