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
Solution:
class Solution {
public:
bool isMatch(const char *s, const char *p) {
if( *p == 0) return *s == 0;
if(*(p+1) != '*'){ //One step match way.
if(*s == *p || (*p) == '.' && (*s) != 0)
{
return isMatch(s+1, p+1);
}
return false;
}
else{ //*,multiple steps match way. while()s++....until(isMatch(s,p+2).
while(*s == *p || ((*p) == '.') && (*s) != 0)
{
if(isMatch(s, p + 2)) //Forward check!!!
{
return true;
}
s++;
}
return isMatch(s, p + 2); //*,multiple steps match function has ended.
}
}
};
if(*(p+1) != '*'){ //One step match way.
if(*s == *p || (*p) == '.' && (*s) != 0)
{
return isMatch(s+1, p+1);
}
return false;
}
else{ //*,multiple steps match way. while()s++....until(isMatch(s,p+2).
while(*s == *p || ((*p) == '.') && (*s) != 0)
{
if(isMatch(s, p + 2)) //Forward check!!!
{
return true;
}
s++;
}
return isMatch(s, p + 2); //*,multiple steps match function has ended.
}
}
};
dp[i][j]表示字串 s[i...len(s)], p[j...len(p)] 是否可以匹配。
那么状态转移方程如下:
dp[i][j] =
c1. p[j+1] != *. if s[i] == p[j] dp[i][j] = dp[i+1][j+1]
else dp[i][j] = false
c2 p[j+1] == '*' (这个情况下,要扩展 *, dp[i][j] 从拓展的情况下,选择一个是真的结果)
if( s[i] == p[j] || p[j] == '.' && (*s) != 0) 当s[i] 和 p[j] 一样的时候,例如 aba, a*b这个时候,i = 0, j = 0, 自然可以匹配a a
如果p[j] == . 因为他可以匹配任何字符,所以和相等关系有基本一样的方式。
并且每一步匹配都要递增 i 的值,如果有成立的,则返回true,否则到匹配终了,返回通配符匹配完成后的结果。
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