LeetCode 10 Regular Expression Matching FaceBook面试题

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

Dynamic Programming Backtracking String

这是一道经典的正则表达式匹配题目，使用暴力法盲目求解会陷入到无穷无尽的边界条件判断中，在这里需要学习动态规划、回溯等处理办法。在这个基础上，才能解决这个问题。

在这里给出一个简洁的递归算法：

bool isMatch(char* s, char* p) {
if (s == NULL || p == NULL) return false;
if (*p == '\0') return *s == '\0';
// ".*" matches "", so we can't check (*s == '\0') here.

if (*(p + 1) == '*')
{
// Here *p != '\0', so this condition equals with
// (*s != '\0' && (*p == '.' || *s == *p)).
while ((*s != '\0' && *p == '.') || *s == *p)
{
if (isMatch(s, p + 2)) return true;
++s;
}

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

return false;
}

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