Careercup - Microsoft面试题 - 4639756264669184

2014-05-12 06:42

题目链接

原题：

Write your own regular expression parser for following condition: 

az*b can match any string that starts with and ends with b and 0 or more Z's between. for e.g. azb, azzzb etc. 

a.b can match anything between a and b e.g. ajsdskjb etc. 

Your function will have to parameters: Input String and Regex. Return true/false if the input string satisfies the regex condition. Note: The input string can contain multiple regex. For e.g. az*bc.g

题目：实现正则表达式中的“*”和“.”功能，不过题目中给定的“.”实际上是“.*”。

解法：Leetcode上有这题，所以我估计是这题的出题者自己记错了，所以说错了“.”的意义。我的Leetcode题解在此：LeetCode - Regular Expression Matching。

代码：

 1 // http://www.careercup.com/question?id=4639756264669184
 2 #include <cstring>
 3 #include <vector>
 4 using namespace std;
 5 
 6 class Solution {
 7 public:
 8     bool isMatch(const char *s, const char *p) {
 9         int i, j;
10         int ls, lp;
11         vector<int> last_i_arr;
12         vector<int> last_j_arr;
13         
14         if (s == nullptr || p == nullptr) {
15             return false;
16         }
17         
18         ls = strlen(s);
19         lp = strlen(p);
20         if (lp == 0) {
21             // empty patterns are regarded as match.
22             return ls == 0;
23         }
24         
25         // validate the pattern string.
26         for (j = 0; j < lp; ++j) {
27             if (p[j] == '*' && (j == 0 || p[j - 1] == '*')) {
28                 // invalid pattern string, can't match.
29                 return false;
30             }
31         }
32         
33         int last_i, last_j;
34         
35         i = j = 0;
36         last_i = -1;
37         last_j = -1;
38         while (i < ls) {
39             if (j + 1 < lp && p[j + 1] == '*') {
40                 last_i_arr.push_back(i);
41                 last_j_arr.push_back(j);
42                 ++last_i;
43                 ++last_j;
44                 j += 2;
45             } else if (p[j] == '.' || s[i] == p[j]) {
46                 ++i;
47                 ++j;
48             } else if (last_j != -1) {
49                 if (p[last_j_arr[last_j]] == '.' || s[last_i_arr[last_i]] == p[last_j_arr[last_j]]) {
50                     // current backtracking position is still available.
51                     i = (++last_i_arr[last_i]);
52                     j = last_j_arr[last_j] + 2;
53                 } else if (last_j > 0) {
54                     while (last_j  >= 0) {
55                         // backtrack to the last backtracking point.
56                         --last_i;
57                         --last_j;
58                         last_i_arr.pop_back();
59                         last_j_arr.pop_back();
60                         if (last_j >= 0 && (p[last_j_arr[last_j]] == '.' || s[last_i_arr[last_i]] == p[last_j_arr[last_j]])) {
61                             i = (++last_i_arr[last_i]);
62                             j = last_j_arr[last_j] + 2;
63                             break;
64                         }
65                     }
66                     if (last_j == -1) {
67                         return false;
68                     }
69                 } else {
70                     // no more backtracking is possible.
71                     return false;
72                 }
73             } else {
74                 return false;
75             }
76         }
77         
78         while (j < lp) {
79             if (j + 1 < lp && p[j + 1] == '*') {
80                 j += 2;
81             } else {
82                 break;
83             }
84         }
85         
86         last_i_arr.clear();
87         last_j_arr.clear();
88         return j == lp;
89     }
90 };

 

 posted on 2014-05-12 06:50  zhuli19901106  阅读(225)  评论(0)  编辑  收藏  举报