Regular Expression Matching(leetcode)
2014-11-24 19:09
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题目:
Implement regular expression matching with support for
题目来源:https://oj.leetcode.com/problems/regular-expression-matching/
解题思路:p当前字符的下一个不是*时,p的当前字符必须和s的当前字符相同,否则就不匹配,返回,若当前字符后面一个是*,则*表示当前字符出现的次数,为0,1,2……,此时就需要把s不断往后移动,知道s当前字符和p的当前字符不相等,把这中间的所有*代表的数字都枚举一遍即可得到结果。
注意:*表示其前面一个字符可以出现的次数!!
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
题目来源:https://oj.leetcode.com/problems/regular-expression-matching/
解题思路:p当前字符的下一个不是*时,p的当前字符必须和s的当前字符相同,否则就不匹配,返回,若当前字符后面一个是*,则*表示当前字符出现的次数,为0,1,2……,此时就需要把s不断往后移动,知道s当前字符和p的当前字符不相等,把这中间的所有*代表的数字都枚举一遍即可得到结果。
注意:*表示其前面一个字符可以出现的次数!!
#include<iostream> using namespace std; bool isMatch(const char *s, const char *p) { if(*p=='\0') return *s=='\0'; if(*(p+1)!='*') { if(*p==*s || (*p=='.'&& *s!='\0'))//一定要注意判断*s!='\0' return isMatch(s+1,p+1); else return false; } else { while(*p==*s || (*p=='.'&& *s!='\0'))//一定要注意判断*s!='\0' { if(isMatch(s,p+2)) return true; s++; } return false; } } int main() { char *s="a",*p=".*..a*"; bool result=isMatch(s,p); system("pause"); return 0; }
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