[LeetCode]044-Wildcard Matching

题目：

‘?’ Matches any single character.

‘*’ Matches any sequence of characters (including the empty sequence).

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”, “*”) → true

isMatch(“aa”, “a*”) → true

isMatch(“ab”, “?*”) → true

isMatch(“aab”, “c*a*b”) → false

Solution(1):

先想到用递归，超时：

bool isMatch(string s, string p)
{
int i;
if(p.size() == 0)
return s.size() == 0?true:false;

if(p[0] == '*')
{
i = 0;
while(p[i] == '*')
i++;
p = p.substr(i);

for(i = 0;i<s.size();i++)
{
if(s[i] == p[0])
{
string m = s.substr(i);
if(isMatch(m,p))
return true;
}
}
//while(s.size()>0)
//{
//  if(isMatch(s,p))
//      return true;
//  s = s.substr(1);
//}
return isMatch(s,p);
}
else if(s.size() > 0 && (s[0] == p[0] || p[0] == '?'))
{
return isMatch(s.substr(1),p.substr(1));
}
return false;
}

Solution(2):

参考C语言版代码后，整理出C++版本。

bool isMatch(string s, string p)
{
int star_index = -1;
int s_index = 0;
int i = 0;
int j = 0;
if(p.size() == 0)
return s.size() == 0?true:false;

while(i < s.size())
{
if(j < p.size() && (p[j] == s[i] || p[j] == '?'))
{
i++;
j++;
continue;
}
if(j < p.size() && p[j] == '*')
{
j++;
star_index = j;
s_index = i;
continue;
}
if(star_index != -1)
{
j = star_index;
s_index ++;
i = s_index;
continue;
}
return false;
}
while(j < p.size() && p[j] == '*')
j++;

return j == p.size()?true:false;
}