LeetCode Problem10 Regular Expression Matching
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'.' Matches any single character.'*' Matches zero or more of the preceding element.The matching should cover the entire input string (not partial).The function prototype should be:bool isMatch(const char *s, const char *p)Some examples:isMatch("aa","a") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true
思路:本题又是一个动态规划的题目,但是不会用找不到状态机。所有本题用的递归来解决的。
在判断s的第i个位置是否匹配时,要先保证s的第i-1个位置的匹配。
本题的难点在于当p中第j个位置出现*时,我们要判断p的第j-1个位置是否等于s的第i个位置
若相等,还要判断s中第i+1,i+2,....的位置是否等于p的第j-1个位置的字符。别的地方就没什么了。
代码如下(已通过leetcode)
public class Solution {public boolean isMatch(String s, String p) {if (s == null || p == null)return false;int lenS = s.length();int lenP = p.length();if (lenP == 0)return lenS == 0;if (lenP == 1) {if (p.equals(s) || p.equals(".") && s.length() == 1) {return true;} elsereturn false;}if (p.charAt(1) == '*') {while (s.length() > 0 && (p.charAt(0) == s.charAt(0) || p.charAt(0) == '.')) {if (isMatch(s, p.substring(2)))return true;s = s.substring(1);}return isMatch(s, p.substring(2));} else {if (s.length() > 0 && (p.charAt(0) == s.charAt(0) || p.charAt(0) == '.')) {return isMatch(s.substring(1), p.substring(1));}return false;}}
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