leetcode Regular Expression Matching
2015-03-07 16:48
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题目:
Implement regular expression matching with support for
解题思路:
思路:递归。根据下一个字符是否是'*'分情况判断。
1.如果p的下一个字符不是'*',只需判断当前字符是否相等,或者p[cur]='.',递归处理s[1]和p[1]。
2.如果是p的下一个'*',则当前s和p相等或者p='.'情况下,依次判断s[0...s.length]和p2是否match。
代码:
参考链接:http://my.oschina.net/jdflyfly/blog/283584
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). The function prototype should be: bool isMatch(const char *s, const char *p) Some examples: isMatch("aa","a") → false isMatch("aa","aa") → true isMatch("aaa","aa") → false isMatch("aa", "a*") → true isMatch("aa", ".*") → true isMatch("ab", ".*") → true isMatch("aab", "c*a*b") → true
解题思路:
思路:递归。根据下一个字符是否是'*'分情况判断。
1.如果p的下一个字符不是'*',只需判断当前字符是否相等,或者p[cur]='.',递归处理s[1]和p[1]。
2.如果是p的下一个'*',则当前s和p相等或者p='.'情况下,依次判断s[0...s.length]和p2是否match。
代码:
public boolean isMatch(String s, String p) { if (s == null) return p == null; if (p == null) return s == null; int lenS = s.length(); int lenP = p.length(); if (lenP == 0) return lenS == 0; if (lenP == 1) { if (s.equals(p) || s.length() == 1 && p.equals(".")) return true; else return false; } if (p.charAt(1) != '*') { if (s.length() > 0 && (s.charAt(0) == p.charAt(0) || p.charAt(0) == '.')) return isMatch(s.substring(1), p.substring(1)); return false; } else { while (s.length() > 0 && (s.charAt(0) == p.charAt(0) || p.charAt(0) == '.')) { if (isMatch(s, p.substring(2))) return true; s = s.substring(1);//s串出现前面字符相同的情况 } return isMatch(s, p.substring(2));//不满足while中的第一个字符相等时,必须比较整个s和p[2]是否相等。如adb和a*b } }
参考链接:http://my.oschina.net/jdflyfly/blog/283584
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