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

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解法一：

這題想法參考了https://oj.leetcode.com/discuss/10133/linear-runtime-and-constant-space-solution，

類似于一種有限狀態機的做法。

主要思想如下：

由于*匹配多少個字符是不一定的，于是首先假設*不匹配任何字符。

當后續匹配過程中出錯，采用回溯的方法，假設*匹配0個字符、1個字符、2個字符……i個字符，然后繼續匹配。

因此s需要有一個spos指針，記錄當p中的*匹配i個字符后，s的重新開始位置。

p也需要一個starpos指針指向*的位置，當回溯過后重新開始時，從starpos的下一位開始。

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;}}} };

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解法二：

模仿Regular Expression Matching的遞歸做法，小數據集上能過。

當*數目太多時會超時。

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;}} };

以下是我的測試代碼，小數據上全部通過：

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; }

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轉載于:https://www.cnblogs.com/ganganloveu/p/4161767.html

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