10. Regular Expression Matching
Given an input string (s
) and a pattern (p
), implement regular expression matching with support for '.'
and '*'
.
'.' Matches any single character. '*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
Note:
s
could be empty and contains only lowercase lettersa-z
.p
could be empty and contains only lowercase lettersa-z
, and characters like.
or*
.
Example 1:
Input: s = "aa" p = "a" Output: false Explanation: "a" does not match the entire string "aa".
Example 2:
Input: s = "aa" p = "a*" Output: true Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
Example 3:
Input: s = "ab" p = ".*" Output: true Explanation: ".*" means "zero or more (*) of any character (.)".
Example 4:
Input: s = "aab" p = "c*a*b" Output: true Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".
Example 5:
Input: s = "mississippi" p = "mis*is*p*." Output: false
class Solution { public boolean isMatch(String text, String pattern) { //递归结束条件 if(pattern.isEmpty()) return text.isEmpty(); //判断 text 是否为空,防止越界,如果 text 为空, 表达式直接判为 false, text.charAt(0)就不会执行了 boolean firstmatch = (!text.isEmpty() && (pattern.charAt(0) == text.charAt(0) || pattern.charAt(0) == '.')); //两种情况 //pattern 直接跳过两个字符。表示 * 前边的字符出现 0 次 //pattern 不变,例如 text = aa ,pattern = a*,第一个 a 匹配,然后 text 的第二个 a 接着和 pattern 的第一个 a 进行匹配。表示 * 用前一个字符替代。 if(pattern.length() >= 2 && pattern.charAt(1) == '*'){ return (isMatch(text, pattern.substring(2)) || firstmatch && isMatch(text.substring(1), pattern)); } else return firstmatch && isMatch(text.substring(1), pattern.substring(1)); } }
https://leetcode.windliang.cc/leetCode-10-Regular-Expression-Matching.html
class Solution { public boolean isMatch(String text, String pattern) { // 多一维的空间,因为求 dp[len - 1][j] 的时候需要知道 dp[len][j] 的情况, // 多一维的话,就可以把 对 dp[len - 1][j] 也写进循环了 boolean[][] dp = new boolean[text.length() + 1][pattern.length() + 1]; // dp[len][len] 代表两个空串是否匹配了,"" 和 "" ,当然是 true 了。 dp[text.length()][pattern.length()] = true; // 从 len 开始减少 for (int i = text.length(); i >= 0; i--) { for (int j = pattern.length(); j >= 0; j--) { // dp[text.length()][pattern.length()] 已经进行了初始化 if(i==text.length()&&j==pattern.length()) continue; boolean first_match = (i < text.length() && j < pattern.length() && (pattern.charAt(j) == text.charAt(i) || pattern.charAt(j) == '.')); if (j + 1 < pattern.length() && pattern.charAt(j + 1) == '*') { dp[i][j] = dp[i][j + 2] || first_match && dp[i + 1][j]; } else { dp[i][j] = first_match && dp[i + 1][j + 1]; } } } return dp[0][0]; } }
dp, 从后往前匹配,和从前往后匹配原则一样。
firstmatch是match第一个字符,接下来判断有✳的时候。题目要求*前面要有东西,所以非dp的情况要先判断pattern是否大于2,dp用了类似的手法