LeetCode Online Judge 题目C# 练习 - 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

        public static bool WildcardMatching(string s, string p)
        {
            return WildcardMathchingHepler(s, 0, p, 0);
        }

        public static bool WildcardMathchingHepler(string s, int i, string p, int j)
        {
            if (i == s.Length)
            {
                if (j == p.Length)
                    return true;
                else if (p[j] == '*')
                    return WildcardMathchingHepler(s, i, p, j + 1);
                else
                    return false;
            }
            else
            {
                if (j == p.Length)
                    return false;
                else if (p[j] == '?' || (s[i] == p[j]))
                    return WildcardMathchingHepler(s, i + 1, p, j + 1);
                else if (p[j] == '*')
                    return WildcardMathchingHepler(s, i, p, j + 1) || WildcardMathchingHepler(s, i + 1, p, j);
                else
                    return false;
            }
        }

代码分析：

　　递归。

　　

 1         public static bool WildcardMatchingDP(string s, string p)
 2         {
 3             if (p.Length == 0 && s.Length == 0)
 4                 return true;
 5             else if (p.Length == 0 && p[0] != '*')
 6                 return false;
 7             else if (s.Length == 0)
 8                 return false;
 9 
10             int countp = 0;
11             for (int i = 0; i < p.Length; i++)
12             {
13                 if (p[i] != '*')
14                     countp++;
15             }
16 
17             if (countp > s.Length)
18                 return false;
19 
20             bool[,] matrix = new bool[p.Length, s.Length];
21             countp = 0;
22             for (int i = 0; i < p.Length; i++)
23             {
24                 if (p[i] != '*')
25                     countp++;
26                 for (int j = 0; j < s.Length; j++)
27                 {
28                     if (i == 0 && j == 0)
29                     {
30                         if (p[i] == '?' || p[i] == '*' || p[i] == s[j])
31                             matrix[i, j] = true;
32                         else
33                             return false;
34                     }
35                     else if (p[i] == '*')
36                     {
37                         if (i - 1 < 0)
38                             matrix[i, j] = true;
39                         else if (j - 1 < 0)
40                         {
41                             if(matrix[i - 1, j] == true)
42                                 matrix[i, j] = true;
43                             else 
44                                 matrix[i, j] = false;
45                         }
46                         else if (matrix[i - 1, j] == true || matrix[i - 1, j - 1] == true || matrix[i, j - 1] == true)
47                             matrix[i, j] = true;
48                         else
49                             matrix[i, j] = false;
50                     }
51                     else if ((p[i] == '?' || p[i] == s[j]) && j + 1 >= countp)
52                     {
53                         if (i - 1 < 0)
54                             matrix[i, j] = false;
55                         else if (j - 1 < 0)
56                         {
57                             if (p[i - 1] == '*' && matrix[i - 1, j] == true)
58                                 matrix[i, j] = true;
59                             else
60                                 matrix[i, j] = false;
61                         }
62                         else if (matrix[i - 1, j - 1] == true && p[i - 1] != '*')
63                             matrix[i, j] = true;
64                         else if (p[i - 1] == '*' && matrix[i - 1, j] == true)
65                         {
66                             if (i - 2 < 0 || j - 1 < 0)
67                                 matrix[i, j] = true;
68                             else
69                                 matrix[i, j] = matrix[i - 2, j - 1];
70                         }
71                         else
72                             matrix[i, j] = false;
73                     }
74                     else
75                     {
76                         matrix[i, j] = false;
77                     }
78                 }
79             }
80 
81             return matrix[p.Length - 1, s.Length - 1];
82         }

代码分析：

　　DP。

posted @ 2012-10-23 04:40  ETCOW  阅读(492)  评论(0)  编辑  收藏  举报