LeetCode（44） Wildcard Matching

题目

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

分析

字符串匹配问题，有两种解决思路：

采用递归实现，但是性能不好，TLE；

贪心解决，参考网址

AC代码

class Solution {
public:
//字符串模式匹配问题，s为匹配串，p为模式串
bool isMatch(string s, string p) {
return isMatch1(s.c_str(), p.c_str());
}

/*方法一：采用贪心的性质  AC*/
bool isMatch1(const char *s, const char *p) {
//? match one
//* match 0,1,2,3..
// aaaabc *c true
const char* star = nullptr;
const char* rs = nullptr;

while (*s) {
if (*s == *p || *p == '?') { //match
s++; p++;
continue;
}
if (*p == '*') {
star = p; // record star
p++; //match from next p
rs = s; // record the position of s , star match 0
continue;
}
if (star != nullptr) { //if have star in front then backtrace
p = star + 1; //reset the position of p
s = rs + 1;
rs++; //star match 1,2,3,4,5....
continue;
}
return false; //if not match return false
}
while (*p == '*') p++; //skip continue star
return *p == '\0'; // successful match
}

/*方法二：利用字符串指针递归解决 性能不好 TLE*/
bool isMatch2(const char *s, const char *p)
{
if (p == NULL || s == NULL)
return false;
else if (*p == '\0')
return (*s == '\0');
else
{
if (*p == '*')
{
while (*p == '*')
++p;

/*处理'*'可以匹配任意长度任意格式的字符串*/
while (*s != '\0')
{
if (isMatch2(s, p))
return true;
++s;
}//while

return isMatch2(s, p);
}
else if((*s!='\0' && *p == '?') || (*s == *p)){
return isMatch2(++s, ++p);
}//else
return false;
}//else
}
};

GitHub测试程序源码