2014.3.1 20:55
Implement regular expression matching with support for '.'
and '*'
.
'.' Matches any single character. '*' Matches zero or more of the preceding element. 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", "a*") → true isMatch("aa", ".*") → true isMatch("ab", ".*") → true isMatch("aab", "c*a*b") → true
Solution:
This problem has some connection with Wildcard Matching, but the backtracking strategy is a bit different.
In this problem, my solution is to match the letters and record the '*'s as they appear.
When a mismatch happened, you keep backtracking until a match is found, while in "Wildcard Matching", you may only backtrack to the last '*'.
If you have no where else to backtrack and there is not a single match for the current letter, return false.
Still, you have to ignore the redundant '*'s at the tail of the pattern string if there are any.
Total time complexity is O(len(s) * len(p)), but it wouldn't appear that large. I guess for more of the test cases it is near O(len(s) + len(p)). Space complexity is O(len(p)), as it is used to record the appearance of '*'s.
Accepted code:
1 // 1RE, 4WA, 1AC, that was quite a problem... no DP and no recursion is tough. 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 };
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