【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 all
                return 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 s
                    p = starpos + 1;
                    s = spos + 1;
                    spos ++;    //let '*' matches one more chars in s
                }
                else
                //*s is too long
                    return 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 match
                else if(starpos != NULL)
                {//maybe '*' matches too few chars in s
                    p = starpos + 1;
                    s = spos + 1;
                    spos ++;    //let '*' matches one more chars in s
                }
                else
                //not match
                    return 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 s
            if(isMatch(s, p+1))
                return true;
                
            //case2: try all possible numbers
            while(*s != 0)
            {
                s ++;
                if(isMatch(s, p+1))
                    return true;
            }
            return false;
        }
        else
        {
            if((*s==*p) || (*p=='?'))
                return isMatch(s+1, p+1);
            else
                return 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;
}