DFS和动态规划——字符串匹配 真蛋疼 为*的情况需考虑匹配0个、1个、2个情况 DFS会超时 正则匹配的话 需要向前看x*的情况 打包处理

192. 通配符匹配

中文

English

判断两个可能包含通配符“？”和“*”的字符串是否匹配。匹配规则如下：

• '?' 可以匹配任何单个字符。
• '*' 可以匹配任意字符串（包括空字符串）。

两个串完全匹配才算匹配成功。

样例

样例1

输入:
"aa"
"a"
输出: false

输出2

输入:
"aa"
"aa"
输出: true

输出3

输入:
"aaa"
"aa"
输出: false

输出4

输入:
"aa"
"*"
输出: true
说明: '*' 可以替换任何字符串

输出5

输入:
"aa"
"a*"
输出: true

样例6

输入:
"ab"
"?*"
输出: true
说明: '?' -> 'a' '*' -> 'b'

样例7

 

?

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class Solution:     """     @param s: A string     @param p: A string includes "?" and "*"     @return: is Match?     """     def isMatch(self, s, p):         # write your code here         m,n = len(s),len(p)         dp = [[False]*(n+1) for i in range(m+1)]                   dp[0][0] = True         for i in range(1, m+1):             dp[i][0] = False                   for j in range(1, n+1):             dp[0][j] = dp[0][j-1] and p[j-1] == '*'                   for i in range(1, m+1):             for j in range(1, n+1):                 if p[j-1] == '*':                     dp[i][j] = dp[i-1][j-1] or dp[i][j-1] or dp[i-1][j]                 else:                     dp[i][j] = dp[i-1][j-1] and (p[j-1] == '?' or p[j-1] == s[i-1])                                       return dp[m][n]                             """         def dfs(s, i, p, j):             if i == len(s) and j == len(p):                 return True                           if i == len(s) and p[j:] == "*":                 return True                           if i == len(s) or j == len(p):                 return False                           if p[j] == '?':                 return dfs(s, i+1, p, j+1)             elif p[j] == '*':                 return dfs(s, i, p, j+1) or dfs(s, i+1, p, j+1) or dfs(s, i+1, p, j)             else:                 if s[i] != p[j]:                     return False                 return dfs(s, i+1, p, j+1)           return dfs(s, 0, p, 0)         """

 

154. 正则表达式匹配

中文

English

实现支持'.'和'*'的正则表达式匹配。

'.'匹配任意一个字母。

'*'匹配零个或者多个前面的元素。

匹配应该覆盖整个输入字符串，而不仅仅是一部分。

需要实现的函数是：bool isMatch(string s, string p)

isMatch("aa","a") → false

isMatch("aa","aa") → true

isMatch("aaa","aa") → false

isMatch("aa", "a*") → true

isMatch("aa", ".*") → true

isMatch("ab", ".*") → true

isMatch("aab", "c*a*b") → true

样例

样例 1:

输入："aa"，"a"
输出：false
解释：
无法匹配

样例 2:

输入："aa"，"a*"
输出：true
解释：
'*' 可以重复 a

?

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class Solution:     """     @param s: A string     @param p: A string includes "." and "*"     @return: A boolean     """       def isMatch(self, s, p):         # write your code here         m, n = len(s), len(p)         dp = [[False] * (n + 1) for i in range(m + 1)]           dp[0][0] = True         for i in range(1, m + 1):             dp[i][0] = False           for j in range(1, n + 1):             if p[j - 1] == '*':                 dp[0][j] = dp[0][j - 1]                 continue               if j < n and p[j - 1] != '*' and p[j] == '*':                 dp[0][j] = dp[0][j - 1]           for i in range(1, m + 1):             for j in range(1, n + 1):                 if p[j - 1] == '*':                     dp[i][j] = dp[i][j - 1]                     continue                   if j < n and p[j] == '*':                     if p[j-1] != '.' and p[j - 1] != s[i - 1]:                         dp[i][j] = dp[i][j - 1]                     else:                         dp[i][j] = dp[i][j - 1] or dp[i - 1][j - 1] or dp[i - 1][j]                 else:                     dp[i][j] = dp[i - 1][j - 1] and (p[j - 1] == '.' or p[j - 1] == s[i - 1])           return dp[m][n]