LeetCode OJ算法题(十):Regular Expression Matching
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题目:
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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("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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