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:
scould be empty and contains only lowercase lettersa-z.pcould 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.