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
题意匹配字符串和正则模式串，只考虑"."和"*"
class Solution {
public:
bool isMatch(const char *s, const char *p) {
int sLen=strlen(s);
int pLen=strlen(p);
if(sLen==0 && pLen==0)return true;
if(sLen!=0 && pLen==0)return false;
if(sLen==0 && pLen==1)return false; //*都是和一个字符成对出现，p长度为1表明不可能再有*
if(p[1]==‘*‘){
if(p[0]==‘.‘){
int k=0; //*匹配的字符数
while(k<=sLen){
if(isMatch(s+k, p+2))return true;
k++;
}
}
else{
int k=0; //*匹配的字符数
while(k<=sLen && s[k]==p[0]){
if(isMatch(s+k, p+2))return true;
k++;
}
if(isMatch(s+k, p+2))return true;
}
}
else{
if(sLen==0)return false; //注意"" "..*"这种特殊的情况
else if(p[0]==‘.‘ || s[0]==p[0])
return isMatch(s+1, p+1);
}
return false;
}
};LeetCode 010 Regular Expression Matching,布布扣,bubuko.com
LeetCode 010 Regular Expression Matching
原文地址：http://blog.csdn.net/harryhuang1990/article/details/25796455
