标签：java leetcode 字符串 regular expression
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
使用递归的思想,分为两种情况：1. p[i + 1] != ‘*‘, 返回 p[i] == s[j] && (s(j + 1), p(i + 1)):递归求解2. p[i + 1] == ‘*‘,分p[i] != s[j], 递归求解（s(j), p(i + 2)）
p[i] == s[j] 递归求解（s(j), p(i + 2)）以及 (s(j + 1) ,p (i + 2))private boolean getMatch(String s, int lenS, int curS, String p, int lenP, int curP) { if (curP == lenP) { return curS == lenS; } if (curP + 1 == lenP || p.charAt(curP + 1) != '*') { if (curS == lenS) { return false; } return (p.charAt(curP) == s.charAt(curS) || p.charAt(curP) == '.') && getMatch(s, lenS, curS + 1, p, lenP, curP + 1); } while (curS < lenS && (s.charAt(curS) == p.charAt(curP) || p.charAt(curP) == '.')) { if (getMatch(s, lenS, curS, p, lenP, curP + 2)) { return true; } curS++; } return getMatch(s, lenS, curS, p, lenP, curP + 2); } public boolean isMatch(String s, String p) { if ((s == null && p == null) || (s.length() == 0 && p.length() == 0)) { return true; } int lenS = s.length(); int lenP = p.length(); return getMatch(s, lenS, 0, p, lenP, 0); }
Regular Expression Matching,布布扣,bubuko.com
标签：java leetcode 字符串 regular expression
原文地址：http://blog.csdn.net/u010378705/article/details/33729511
