題目闡釋：

正則匹配字符串，用程序實現

關鍵理解：

正則匹配，動態規劃思想，一個個向后追溯，后面的依賴前面的匹配成功。 正則和待匹配的字符串長度不一，統一到正則字符串的index索引上，每次的字符串index移動，都以匹配到的正則的index為準。 正則由于*?的存在，所以有多種狀態，中間狀態儲存都需要記錄下來。然后以這些狀態為動態的中轉，繼續判斷到最后。 最后正則匹配字符串是否成功的判斷依據，就是正則字符串的最大index，是否出現在遍歷到最后的狀態列表中。

錯誤之處：

多處動態變化，導致無法入手，*沒有處理思路，沒有找到匹配成功的條件

應用：

正則屬于多條路徑問題，可以推理到 多種渠道的問題，匹配成功當前的才往后推 *相當于無限向后匹配，所以無限循環使用，看能否匹配成功。

Wildcard Matching

Given an input string (s) and a pattern (p), 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). Note: s could be empty and contains only lowercase letters a-z. p could be empty and contains only lowercase letters a-z, and characters like ? or *. Example 1:

Input:

s = "aa" p = "a" Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:

s = "aa" p = "*" Output: true
Explanation: '*' matches any sequence.

Example 3:

Input:

s = "cb" p = "?a" Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.

Example 4:

Input:

s = "adceb" p = "*a*b" Output: true
Explanation: The first '' matches the empty sequence, while the second '' matches the substring "dce".

Example 5:

Input:

s = "acdcb" p = "a*c?b" Output: false class Solution(object):def isMatch(self, s, p):""":type s: str:type p: str:rtype: bool"""transfer = {}index=0for char in p:if char=='*':transfer[index,char]=indexelse:transfer[index,char]=index+1index+=1accept=index# index=0state = {0}for char in s:state_tmp=set()for index in state:for char_prob in [char,'?','*']:index_next=transfer.get((index,char_prob))state_tmp.add(index_next)state=state_tmpreturn accept in stateif __name__=='__main__':s = "acdcb"p = "a*c?b"p = "a**c?d"st=Solution()out=st.isMatch(s,p)print(out)

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