[leetcode] Regular Expression Matching

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

»
Solve this problem

待匹配串为S，匹配串为P。

简化问题，假设P中没有'*'，那么我们只需要一位一位的比较S和P。

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for (int i = 0; i < length; i++) {

if (S[i] != P[i] && P[i] != '.') {

匹配失败;

}

}

匹配成功;

回到原问题，对于S[i]和P[j]：

如果P[j+1]!='*'，S[i] == P[j]=>匹配下一位(i+1, j+1)，S[i]!=P[j]=>匹配失败；

如果P[j+1]=='*'，S[i]==P[j]=>匹配下一位(i+1, j+2)或者(i, j+2)，S[i]!=P[j]=>匹配下一位(i,j+2)。

匹配成功的条件为S[i]=='\0' && P[j]=='\0'。

采用递归实现：

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class Solution {

public:

bool isMatch(const char *s, const char *p) {

// Start typing your C/C++ solution below

// DO NOT write int main() function

if (*p == '\0') {

return *s == '\0';

}

if (*(p + 1) == '*') {

while (*s != '\0' && (*s == *p || *p == '.')) {

if (isMatch(s, p + 2)) {

return true;

}

s++;

}

return isMatch(s, p + 2);

}

else {

if (*s != '\0' && (*s == *p || *p == '.')) {

return isMatch(s + 1, p + 1);

}

}

return false;

}

};