Wildcard Matching
Implement wildcard pattern matching with support for ‘?‘
and ‘*‘
.
‘?‘ 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
思路:本题是正常的通配符匹配,可以使用循环,也可以使用递归。使用循环时,每次遇到‘*‘时,要记录他的位置,这样当匹配失败时,返回到该位置重新匹配。
class Solution {
public:
bool isMatch(const char *s, const char *p) {
const char* sBegin = NULL,*pBegin = NULL;
while(*s)
{
if(*s == *p || *p == '?')
{
++s;
++p;
}
else if(*p == '*')
{
pBegin = p;//记录通配符的位置
sBegin = s;
++p;
}
else if(pBegin != NULL)
{
p = pBegin + 1;//重通配符的下一个字符开始
++sBegin;//每次多统配一个
s = sBegin;
}
else return false;
}
while(*p == '*')++p;
return (*p == '\0');
}
};
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
思路:本题和上面不同之处在于,此时的通配符‘*‘是代表0到多个前一个字符,而不是任一个字符。所以,当下一个字符是‘*‘时,如果当前字符相等,则反复跳过当前字符去匹配‘*‘后面的字符,如果不相等,则直接匹配‘*‘后面的字符。
class Solution { public: bool isMatch(const char *s, const char *p) { if(*p == '\0')return *s == '\0'; if(*(p+1) != '*') { if(*s != '\0' && (*s == *p || *p == '.'))return isMatch(s+1,p+1); } else { //s向后移动0、1、2……分别和p+2进行匹配 while(*s != '\0' && (*s == *p || *p == '.')) { if(isMatch(s,p+2))return true; ++s; } return isMatch(s,p+2); } return false; } };
leetcode之通配符
时间: 2024-10-12 08:37:37