Regular Expression Matching

Description:

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

解题思路：

题意为实现正则表达式的匹配。其中支持.和*

分成三种情况。设当前字符串和模式的下标分别为i，j

1、若当前模式匹配完毕，即p[j]==’\0’，若字符串结束，则返回true，否则返回false

2、若模式的下一个字符不是，即p[j+1]!=’‘。这里分情况讨论。

(1) 若s[i]==p[j]，递归验证i+1, j+1

(2) 若p[i]==’.’且s[i]!=’\0’，递归验证i+1, j+1

(3) 否则，返回false

3、若模式的下一个字符是，即p[j+1]==’‘，则不断通过递归回溯i+k，j+2（k从0增长至len(s)-i，j+2意味着越过当前字符和*）。

代码：

#include <iostream>

using namespace std;

bool isMatch(const char *s, const char *p)
{
if (*p == '\0')
return *s == '\0';

if (*(p+1) != '*')
{
if ((*s == *p) || (*p == '.' && *s != '/0'))
return isMatch(s+1, p+1);
else
return false;
}
else
{
while((*s == *p) || (*p == '.' && *s != '/0'))
{
if(isMatch(s, p+2))
return true;
s++;
}

return isMatch(s, p+2);
}
}

int main()
{
char *s1 = "abbbc", *p1 = "ab*c";
char *s2 = "ac", *p2 = "ab*c";
char *s3 = "abbc", *p3 = "ab*bbc";
char *s4 = "abc", *p4 = "abbc";

cout<<boolalpha<<isMatch(s1, p1)<<endl;
cout<<boolalpha<<isMatch(s2, p2)<<endl;
cout<<boolalpha<<isMatch(s3, p3)<<endl;
cout<<boolalpha<<isMatch(s4, p4)<<endl;

return 0;
}

测试：



参考链接：

http://www.2cto.com/kf/201404/290845.html

http://www.2cto.com/kf/201506/410752.html

http://www.acmerblog.com/leetcode-solution-regular-expression-matching-6221.html