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 lettersa-z
.p
could be empty and contains only lowercase lettersa-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
AC code:
class Solution { public: bool isMatch(string s, string p) { int len1 = s.length(), len2 = p.length(); // 初始化 vector<vector<bool> > dp(len1+1, vector<bool>(len2+1, false)); dp[0][0] = true; // "a" // "aa*" ---> "a" for (int i = 1; i <= len2; ++i) { if (p[i-1] == '*' && dp[0][i-2]) dp[0][i] = true; } // DP for (int i = 1; i <= len1; ++i) { for (int j = 1; j <= len2; ++j) { if (s[i-1] == p[j-1] || p[j-1] == '.') { dp[i][j] = dp[i-1][j-1]; } else if (p[j-1] == '*') { if (p[j-2] != '.' && p[j-2] != s[i-1]) dp[i][j] = dp[i][j-2]; else { dp[i][j] = (dp[i][j-2] || dp[i-1][j] || dp[i][j-1]); } } } } return dp[len1][len2]; } };
Runtime: 12 ms, faster than 51.36% of C++ online submissions for Regular Expression Matching.