19.1.25 [LeetCode10]Regular Expression Matching

Given an input string (s) and a pattern (p), 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).

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 = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

题解

一开始搞了个dp，很慢：

 1 class Solution {
 2 public:
 3     bool match[1005][1005];
 4     bool equal(char a, char b) {
 5         if (a == '.')return true;
 6         else return a == b;
 7     }
 8     bool isMatch(string s, string p) {
 9         memset(match, 0, sizeof(match));
10         string _p = p;
11         bool isstar[1005] = { false };
12         int pt = 0;
13         for (int i = 0; p[i]; i++) {
14             _p[pt++] = p[i];
15             if (p[i + 1] == '*') {
16                 i++;
17                 isstar[pt - 1] = true;
18             }
19         }
20         match[0][0] = true;
21         for(int i=0;isstar[i];i++)match[i+1][0] = true;
22         for(int i=1;i<=pt;i++)
23             for (int j=1; s[j-1]; j++) {
24                 for (int k = 1; k <= i; k++) {
25                     if (match[k][j])continue;
26                     if (match[k - 1][j - 1] && equal(_p[k - 1], s[j - 1])) {
27                         match[k][j] = true;
28                         for (int t = k+1; isstar[t - 1]; t++)
29                             match[t][j] = true;
30                     }
31                 }
32                 if (match[i][j - 1] && isstar[i - 1] && equal(_p[i - 1], s[j - 1])) {
33                     match[i][j] = true;
34                     for (int t = i + 1; isstar[t - 1]; t++)
35                         match[t][j] = true;
36                 }
37             }
38         return match[pt][s.size()];
39     }
40 };

View Code

后来弄了个递归果然更慢了

 1 class Solution {
 2 public:
 3     bool isMatch(string s,string p) {
 4         if (p == "")return s == "";
 5         if (s == ""&&p.size() <= 1)return false;
 6         if (s == ""&&p[1] != '*')return false;
 7         if (p.size()>=2&&p[1] == '*') {
 8             int i, l = s.size();
 9             bool ans = isMatch(s, p.substr(2, p.size() - 2));
10             for (i = 0; (s[i] == p[0]|| p[0] == '.') && i < l; i++)
11                 ans=ans||isMatch(s.substr(i+1,s.size()-i-1), p.substr(2,p.size()-2));
12             return ans;
13         }
14         if(p[0] == '.'||p[0] == s[0])return isMatch(s.substr(1, s.size() - 1), p.substr(1, p.size() - 1));
15         return false;
16     }
17 };

View Code

唉……好吧，又回到dp……是刚刚那个dp太混乱了

 1 class Solution {
 2 public:
 3     bool isMatch(string s,string p) {
 4         int m = s.size(), n = p.size();
 5         vector<vector<bool>>dp(m + 1, vector<bool>(n + 1, false));
 6         dp[0][0] = true;
 7         for (int i = 0; i <= m; i++) {
 8             for (int j = 1; j <= n; j++) {
 9                 if (j > 1 && p[j - 1] == '*')
10                     dp[i][j] = dp[i][j - 2] ||
11                     (i > 0 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') && dp[i - 1][j]);
12                 else
13                     dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
14             }
15         }
16         return dp[m][n];
17     }
18 };

View Code