10 Regular Expression Matching

10. Regular Expression Matching

1. Question

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 theentireinput 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

2. Implementation

(1) Recursion

思路: 这题要分清况讨论

• 如果p是空，则查看s是否为空

• 如果p长度大于1且p[1] == *, 因为这里的*必须匹配前面的字符0次或多次

  1. 如果匹配0次的话，就继续递归地比较s和p.substring(2)

     2.如果匹配多次，则需要s[0] == p[0] 或者 p[0] == '.'， 然后递归地比较s.substring(1)和p

• 如果上述情况都不成立的话，则在s[0] == p[0] 或者 p[0] == '.'的情况下，递归比较s.substring(1)和p.substring(1)

class Solution {
    public boolean isMatch(String s, String p) {
        if (p.length() == 0) {
            return s.length() == 0;
        }

        if (p.length() > 1 && p.charAt(1) == '*') {
            return isMatch(s, p.substring(2)) || (s.length() != 0 && (s.charAt(0) == p.charAt(0) || p.charAt(0) == '.') && isMatch(s.substring(1), p));
        }
        else {
            return s.length() != 0 && (s.charAt(0) == p.charAt(0) || p.charAt(0) == '.') && isMatch(s.substring(1), p.substring(1));
        }
    }
}

（2）DP

思路:

1. P[i][j] = P[i - 1][j - 1], if p[j - 1] != '*' && (s[i - 1] == p[j - 1] || p[j - 1] == '.');

2. P[i][j] = P[i][j - 2], if p[j - 1] == '*' and the pattern repeats for 0 times;

3. P[i][j] = P[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'), if p[j - 1] == '*' and the pattern repeats for at least 1 times.

class Solution {
    public boolean isMatch(String s, String p) {
        int m = s.length();
        int n = p.length();
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;

        // s要从0开始，因为s的开头可以被*看为空，如果*表示不匹配之前的character
        for (int i = 0; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (j > 1  && p.charAt(j - 1) == '*') {
                    dp[i][j] = dp[i][j - 2] || (i > 0 && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') && dp[i - 1][j]);
                }
                else {
                    dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.');
                }
            }
        }
        return dp[m][n];
    }
}

3. Time & Space Complexity

Recursion: 时间复杂度O(2^n), 因为，假设P全是a*a*a*这样组成，s = aaaaaaaa 而s的每一个字符都有2种可能：与当前的a*匹配,或者与下一个a*匹配（前一个匹配空), 这样假设s有n个字符，则实际上的复杂度是2^n, 空间复杂度O(L), L是s的长度