publicbooleanisMatch(String text, String pattern){boolean[][] dp =newboolean[text.length() + 1][pattern.length() + 1]; dp[text.length()][pattern.length()]=true;for(inti = text.length(); i >= 0; i--){for(intj = pattern.length() - 1; j>= 0; j--){booleanfirst_match = ...
}privatebooleanmatch(String s,String p,inti,intj){//其中i,j分别为开始下标if(j==p.length())//匹配至长度相同returni==s.length();if(j==p.length()-1 || p.charAt(j+1)!='*'){//匹配至下一个字符不为'*'if(i==s.length() || s.charAt(i)!=p.charAt(j) && p.charAt(j)!='....
match(text[1:], pattern) 以上这两条路是并行执行的,而且只要有一条路满足就可以,所以要用 or 连接。 如果上述条件不满足,就可以认为pattern[0]和pattern[1]里面均不包含*(至少在当前 pattern 的前两个位置是没有*的,后面的位置有*的情况先不管。因为代码是递归进行的,所以后面的位置如果有*,早晚有一天这...
Memory Usage: 9.1 MB, less than 0.94% of C++ online submissions for Regular Expression Matching. Best Solution(8ms) classSolution{public:boolisMatch(strings,stringp){booldp[1005][1005];dp[0][0]=true;dp[0][1]=false;intlen1=s.size(),len2=p.size();for(intj=1;j<len2;j+=2)dp[0]...
def isMatch(self, s: str, p: str) -> bool: if not p: if not s: return True return False if not s: if len(p)>1 and p[1]=='*': return self.isMatch(s,p[2:]) return False if len(p) == 1: if len(s) == 1 and (p[0] == '.' or p[0] == s[0]): return...
'*' 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 ...
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Consider the regular expression a?nan. It matches the string an when the a? are chosen not to match any letters, leaving the entire string to be matched by the an. Backtracking regular expression implementations implement the zero-or-one ? by first trying one and then zero. There are n ...
return re.search('^' + p + '$',s)!=None 1. 2. 3. 4. 5. 6. 7. 8. 9. '^'表示匹配字符串开头,'$'表示匹配字符串结尾。 re.match只匹配字符串的开始,如果字符串开始不符合正则表达式,则匹配失败,函数返回None;而re.search匹配整个字符串,直到找到一个匹配。
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