LeetCode OJ算法题(十):Regular Expression Matching
2014-07-15 14:06
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题目:
Implement regular expression matching with support for
解法:
本题旨在判断两个字符串是否完全一样,关键在于*号的支持,*号表示前面的字符重复0次或者更多次,因此需要对每个字符后面是否为*号加以区分,若为*号,则该符号可能重复0次~
下面分情况讨论(当前指针处于si,pi):
1、若pi+1不为*,则直接比较si和pi对应的字符是否相等,如果不等直接false,如果相等,则继续比较si+1和pi+1状态;
2、若pi+1为*,那么我们可以把s字符串记作{Xi},p字符串记作{Z*Yj},那么我们需要依次比较{Xi}vs{Yj},{Xi}vsZ{Yj},{Xi}vsZZ{Yj}。。。。。。等,其中,当我们比较{Xi}与Z...Z(n个Z){Yj}时,必有{Xi}的前n个字符都是Z,循环终止的条件就是n大于{Xi}中前缀Z的个数。
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
解法:
本题旨在判断两个字符串是否完全一样,关键在于*号的支持,*号表示前面的字符重复0次或者更多次,因此需要对每个字符后面是否为*号加以区分,若为*号,则该符号可能重复0次~
下面分情况讨论(当前指针处于si,pi):
1、若pi+1不为*,则直接比较si和pi对应的字符是否相等,如果不等直接false,如果相等,则继续比较si+1和pi+1状态;
2、若pi+1为*,那么我们可以把s字符串记作{Xi},p字符串记作{Z*Yj},那么我们需要依次比较{Xi}vs{Yj},{Xi}vsZ{Yj},{Xi}vsZZ{Yj}。。。。。。等,其中,当我们比较{Xi}与Z...Z(n个Z){Yj}时,必有{Xi}的前n个字符都是Z,循环终止的条件就是n大于{Xi}中前缀Z的个数。
public class No10_RegularExpressionMatching { public static void main(String[] args){ System.out.println(isMatch("ca", "c*a")); } public static boolean isMatch(String s, String p) { if(p.length() == 0) return s.length() == 0; if(p.length() == 1) return s.length() == 1 && (s.equals(p) || p.charAt(0) == '.'); int si = 0, pi = 0; if(p.charAt(pi+1) != '*'){ if(si<s.length() && (s.charAt(si) == p.charAt(pi) || p.charAt(pi) == '.')){ return isMatch(s.substring(si+1), p.substring(pi+1)); } return false; } else{ while(si<s.length() && (s.charAt(si) == p.charAt(pi) || p.charAt(pi) == '.')){ if(isMatch(s.substring(si), p.substring(pi+2))) return true; si++; } return isMatch(s.substring(si), p.substring(pi+2)); } } }
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