【LeetCode】44. Wildcard Matching (2 solutions)

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

 

解法一：

这题想法参考了https://oj.leetcode.com/discuss/10133/linear-runtime-and-constant-space-solution，

类似于一种有限状态机的做法。

主要思想如下：

由于*匹配多少个字符是不一定的，于是首先假设*不匹配任何字符。

当后续匹配过程中出错，采用回溯的方法，假设*匹配0个字符、1个字符、2个字符……i个字符，然后继续匹配。

因此s需要有一个spos指针，记录当p中的*匹配i个字符后，s的重新开始位置。

p也需要一个starpos指针指向*的位置，当回溯过后重新开始时，从starpos的下一位开始。

class Solution {
public:bool isMatch(const char *s, const char *p) {char* starpos = NULL;char* spos = NULL;while(true){if(*s == 0 && *p == 0)//match allreturn true;else if(*s == 0 && *p != 0){//successive *p must be all '*'while(*p != 0){if(*p != '*')return false;p ++;}return true;}else if(*s != 0 && *p == 0){if(starpos != NULL){//maybe '*' matches too few chars in sp = starpos + 1;s = spos + 1;spos ++;    //let '*' matches one more chars in s
                }else//*s is too longreturn false;}else{if(*p == '?' || *s == *p){s ++;p ++;}else if(*p == '*'){starpos = (char*)p;spos = (char*)s;p ++;   //start successive matching from "'*' matches non"
                }//currently not matchelse if(starpos != NULL){//maybe '*' matches too few chars in sp = starpos + 1;s = spos + 1;spos ++;    //let '*' matches one more chars in s
                }else//not matchreturn false;}}}
};

 

解法二：

模仿Regular Expression Matching的递归做法，小数据集上能过。

当*数目太多时会超时。

class Solution {
public:bool isMatch(const char *s, const char *p) {if(*p == 0)return (*s == 0);if(*p == '*'){//case1: *p matches 0 chars in sif(isMatch(s, p+1))return true;//case2: try all possible numberswhile(*s != 0){s ++;if(isMatch(s, p+1))return true;}return false;}else{if((*s==*p) || (*p=='?'))return isMatch(s+1, p+1);elsereturn false;}}
};

以下是我的测试代码，小数据上全部通过：

int main()
{Solution s;cout << s.isMatch("aa","a") << endl;cout << s.isMatch("aa","aa") << endl;cout << s.isMatch("aaa","aa") << endl;cout << s.isMatch("aa","*") << endl;cout << s.isMatch("aa","a*") << endl;cout << s.isMatch("ab","?*") << endl;cout << s.isMatch("aab","c*a*b") << endl;return 0;
}