leetcode之萬用字元

標籤：wildcard matching   regular expression m   leetcode   wildcard   pattern matching   

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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false

思路：本題是正常的萬用字元匹配，可以使用迴圈，也可以使用遞迴。使用迴圈時，每次遇到‘*‘時，要記錄他的位置，這樣當匹配失敗時，返回到該位置重新匹配。

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;//記錄萬用字元的位置    sBegin = s;    ++p;    }    else if(pBegin != NULL)    {    p = pBegin + 1;//重萬用字元的下一個字元開始    ++sBegin;//每次多統配一個    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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

思路：本題和上面不同之處在於，此時的萬用字元‘*‘是代表0到多個前一個字元，而不是任一個字元。所以，當下一個字元是‘*‘時，如果當前字元相等，則反覆跳過當前字元去匹配‘*‘後面的字元，如果不相等，則直接匹配‘*‘後面的字元。

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向後移動0、1、2……分別和p+2進行匹配    while(*s != '\0' && (*s == *p || *p == '.'))    {    if(isMatch(s,p+2))return true;    ++s;    }    return isMatch(s,p+2);    }    return false;    }};