[LeetCode] 10. Regular Expression Matching 正则表达式匹配

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

通配符匹配问题，和44. Wildcard Matching类似。'.'表示1个字符，'*'表示它前面的字符可以有0个，1个或是多个，比如：字符串a*b，可以表示b或是aaab，即a的个数可以是0个，也可以是多个。

解法1：DP

解法2: 递归

Java:

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public boolean isMatch(String s, String p) {
 
    if (s == null || p == null) {
        return false;
    }
    boolean[][] dp = new boolean[s.length()+1][p.length()+1];
    dp[0][0] = true;
    for (int i = 0; i < p.length(); i++) {
        if (p.charAt(i) == '*' && dp[0][i-1]) {
            dp[0][i+1] = true;
        }
    }
    for (int i = 0 ; i < s.length(); i++) {
        for (int j = 0; j < p.length(); j++) {
            if (p.charAt(j) == '.') {
                dp[i+1][j+1] = dp[i][j];
            }
            if (p.charAt(j) == s.charAt(i)) {
                dp[i+1][j+1] = dp[i][j];
            }
            if (p.charAt(j) == '*') {
                if (p.charAt(j-1) != s.charAt(i) && p.charAt(j-1) != '.') {
                    dp[i+1][j+1] = dp[i+1][j-1];
                } else {
                    dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1]);
                }
            }
        }
    }
    return dp[s.length()][p.length()];
}

　　

Java:

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public boolean isMatch(String s, String p) {
    // base case
    if (p.length() == 0) {
        return s.length() == 0;
    }
  
    // special case
    if (p.length() == 1) {
  
        // if the length of s is 0, return false
        if (s.length() < 1) {
            return false;
        }
  
        //if the first does not match, return false
        else if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != '.')) {
            return false;
        }
  
        // otherwise, compare the rest of the string of s and p.
        else {
            return isMatch(s.substring(1), p.substring(1));
        }
    }
  
    // case 1: when the second char of p is not '*'
    if (p.charAt(1) != '*') {
        if (s.length() < 1) {
            return false;
        }
        if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != '.')) {
            return false;
        } else {
            return isMatch(s.substring(1), p.substring(1));
        }
    }
  
    // case 2: when the second char of p is '*', complex case.
    else {
        //case 2.1: a char & '*' can stand for 0 element
        if (isMatch(s, p.substring(2))) {
            return true;
        }
  
        //case 2.2: a char & '*' can stand for 1 or more preceding element,
        //so try every sub string
        int i = 0;
        while (i<s.length() && (s.charAt(i)==p.charAt(0) || p.charAt(0)=='.')){
            if (isMatch(s.substring(i + 1), p.substring(2))) {
                return true;
            }
            i++;
        }
        return false;
    }
}

Java:

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public class Solution {
    public boolean isMatch(String s, String p) {
  
        if(p.length() == 0)
            return s.length() == 0;
  
        //p's length 1 is special case   
        if(p.length() == 1 || p.charAt(1) != '*'){
            if(s.length() < 1 || (p.charAt(0) != '.' && s.charAt(0) != p.charAt(0)))
                return false;
            return isMatch(s.substring(1), p.substring(1));   
  
        }else{
            int len = s.length();
  
            int i = -1;
            while(i<len && (i < 0 || p.charAt(0) == '.' || p.charAt(0) == s.charAt(i))){
                if(isMatch(s.substring(i+1), p.substring(2)))
                    return true;
                i++;
            }
            return false;
        }
    }
}

Python:　

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# dp with rolling window
class Solution:
    # @return a boolean
    def isMatch(self, s, p):
        k = 3
        result = [[False for j in xrange(len(p) + 1)] for i in xrange(k)]
 
        result[0][0] = True
        for i in xrange(2, len(p) + 1):
            if p[i-1] == '*':
                result[0][i] = result[0][i-2]
 
        for i in xrange(1,len(s) + 1):
            if i > 1:
                result[0][0] = False
            for j in xrange(1, len(p) + 1):
                if p[j-1] != '*':
                    result[i % k][j] = result[(i-1) % k][j-1] and (s[i-1] == p[j-1] or p[j-1] == '.')
                else:
                    result[i % k][j] = result[i % k][j-2] or (result[(i-1) % k][j] and (s[i-1] == p[j-2] or p[j-2] == '.'))
 
        return result[len(s) % k][len(p)]

Python:

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# dp
# Time:  O(m * n)
# Space: O(m * n)
class Solution2:
    # @return a boolean
    def isMatch(self, s, p):
        result = [[False for j in xrange(len(p) + 1)] for i in xrange(len(s) + 1)]
 
        result[0][0] = True
        for i in xrange(2, len(p) + 1):
            if p[i-1] == '*':
                result[0][i] = result[0][i-2]
 
        for i in xrange(1,len(s) + 1):
            for j in xrange(1, len(p) + 1):
                if p[j-1] != '*':
                    result[i][j] = result[i-1][j-1] and (s[i-1] == p[j-1] or p[j-1] == '.')
                else:
                    result[i][j] = result[i][j-2] or (result[i-1][j] and (s[i-1] == p[j-2] or p[j-2] == '.'))
 
        return result[len(s)][len(p)]

Python:

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# iteration
class Solution3:
    # @return a boolean
    def isMatch(self, s, p):
        p_ptr, s_ptr, last_s_ptr, last_p_ptr = 0, 0, -1, -1
        last_ptr = []
        while s_ptr < len(s):
            if p_ptr < len(p) and (p_ptr == len(p) - 1 or p[p_ptr + 1] != '*') and \
            (s_ptr < len(s) and (p[p_ptr] == s[s_ptr] or p[p_ptr] == '.')):
                    s_ptr += 1
                    p_ptr += 1
            elif p_ptr < len(p) - 1 and (p_ptr != len(p) - 1 and p[p_ptr + 1] == '*'):
                p_ptr += 2
                last_ptr.append([s_ptr, p_ptr])
            elif  last_ptr:
                [last_s_ptr, last_p_ptr] = last_ptr.pop()
                while last_ptr and p[last_p_ptr - 2] != s[last_s_ptr] and p[last_p_ptr - 2] != '.':
                    [last_s_ptr, last_p_ptr] = last_ptr.pop()
 
                if p[last_p_ptr - 2] == s[last_s_ptr] or p[last_p_ptr - 2] == '.':
                    last_s_ptr += 1
                    s_ptr = last_s_ptr
                    p_ptr = last_p_ptr
                    last_ptr.append([s_ptr, p_ptr])
                else:
                    return False
            else:
                return False
 
        while p_ptr < len(p) - 1 and p[p_ptr] == '.' and p[p_ptr + 1] == '*':
            p_ptr += 2
 
        return p_ptr == len(p)

Python:　　

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# recursive
class Solution4:
    # @return a boolean
    def isMatch(self, s, p):
        if not p:
            return not s
 
        if len(p) == 1 or p[1] != '*':
            if len(s) > 0 and (p[0] == s[0] or p[0] == '.'):
                return self.isMatch(s[1:], p[1:])
            else:
                return False
        else:
            while len(s) > 0 and (p[0] == s[0] or p[0] == '.'):
                if self.isMatch(s, p[2:]):
                    return True
                s = s[1:]
            return self.isMatch(s, p[2:])

Python:

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class Solution(object):
    def isMatch(self, s, p):
        # The DP table and the string s and p use the same indexes i and j, but
        # table[i][j] means the match status between p[:i] and s[:j], i.e.
        # table[0][0] means the match status of two empty strings, and
        # table[1][1] means the match status of p[0] and s[0]. Therefore, when
        # refering to the i-th and the j-th characters of p and s for updating
        # table[i][j], we use p[i - 1] and s[j - 1].
 
        # Initialize the table with False. The first row is satisfied.
        table = [[False] * (len(s) + 1) for _ in range(len(p) + 1)]
 
        # Update the corner case of matching two empty strings.
        table[0][0] = True
 
        # Update the corner case of when s is an empty string but p is not.
        # Since each '*' can eliminate the charter before it, the table is
        # vertically updated by the one before previous. [test_symbol_0]
        for i in range(2, len(p) + 1):
            table[i][0] = table[i - 2][0] and p[i - 1] == '*'
 
        for i in range(1, len(p) + 1):
            for j in range(1, len(s) + 1):
                if p[i - 1] != "*":
                    # Update the table by referring the diagonal element.
                    table[i][j] = table[i - 1][j - 1] and \
                                  (p[i - 1] == s[j - 1] or p[i - 1] == '.')
                else:
                    # Eliminations (referring to the vertical element)
                    # Either refer to the one before previous or the previous.
                    # I.e. * eliminate the previous or count the previous.
                    # [test_symbol_1]
                    table[i][j] = table[i - 2][j] or table[i - 1][j]
 
                    # Propagations (referring to the horizontal element)
                    # If p's previous one is equal to the current s, with
                    # helps of *, the status can be propagated from the left.
                    # [test_symbol_2]
                    if p[i - 2] == s[j - 1] or p[i - 2] == '.':
                        table[i][j] |= table[i][j - 1]
 
        return table[-1][-1]
 
 
class TestSolution(unittest.TestCase):
    def test_none_0(self):
        s = ""
        p = ""
        self.assertTrue(Solution().isMatch(s, p))
 
    def test_none_1(self):
        s = ""
        p = "a"
        self.assertFalse(Solution().isMatch(s, p))
 
    def test_no_symbol_equal(self):
        s = "abcd"
        p = "abcd"
        self.assertTrue(Solution().isMatch(s, p))
 
    def test_no_symbol_not_equal_0(self):
        s = "abcd"
        p = "efgh"
        self.assertFalse(Solution().isMatch(s, p))
 
    def test_no_symbol_not_equal_1(self):
        s = "ab"
        p = "abb"
        self.assertFalse(Solution().isMatch(s, p))
 
    def test_symbol_0(self):
        s = ""
        p = "a*"
        self.assertTrue(Solution().isMatch(s, p))
 
    def test_symbol_1(self):
        s = "a"
        p = "ab*"
        self.assertTrue(Solution().isMatch(s, p))
 
    def test_symbol_2(self):
        # E.g.
        #   s a b b
        # p 1 0 0 0
        # a 0 1 0 0
        # b 0 0 1 0
        # * 0 1 1 1
        s = "abb"
        p = "ab*"
        self.assertTrue(Solution().isMatch(s, p))
 
 
if __name__ == "__main__":
    unittest.main()　　

C++:　　

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class Solution {
public:
    bool isMatch(string s, string p) {
        if (p.empty())    return s.empty();
         
        if ('*' == p[1])
            // x* matches empty string or at least one character: x* -> xx*
            // *s is to ensure s is non-empty
            return (isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p));
        else
            return !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p.substr(1));
    }
};
 
class Solution {
public:
    bool isMatch(string s, string p) {
        /**
         * f[i][j]: if s[0..i-1] matches p[0..j-1]
         * if p[j - 1] != '*'
         *      f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1]
         * if p[j - 1] == '*', denote p[j - 2] with x
         *      f[i][j] is true iff any of the following is true
         *      1) "x*" repeats 0 time and matches empty: f[i][j - 2]
         *      2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j]
         * '.' matches any single character
         */
        int m = s.size(), n = p.size();
        vector<vector<bool>> f(m + 1, vector<bool>(n + 1, false));
         
        f[0][0] = true;
        for (int i = 1; i <= m; i++)
            f[i][0] = false;
        // p[0.., j - 3, j - 2, j - 1] matches empty iff p[j - 1] is '*' and p[0..j - 3] matches empty
        for (int j = 1; j <= n; j++)
            f[0][j] = j > 1 && '*' == p[j - 1] && f[0][j - 2];
         
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++)
                if (p[j - 1] != '*')
                    f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]);
                else
                    // p[0] cannot be '*' so no need to check "j > 1" here
                    f[i][j] = f[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && f[i - 1][j];
         
        return f[m][n];
    }
};

C++:

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class Solution {
public:
    bool isMatch(string s, string p) {
        if (p.empty())    return s.empty();
         
        if ('*' == p[1])
            // x* matches empty string or at least one character: x* -> xx*
            // *s is to ensure s is non-empty
            return (isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p));
        else
            return !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p.substr(1));
    }
};
 
class Solution {
public:
    bool isMatch(string s, string p) {
        /**
         * f[i][j]: if s[0..i-1] matches p[0..j-1]
         * if p[j - 1] != '*'
         *      f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1]
         * if p[j - 1] == '*', denote p[j - 2] with x
         *      f[i][j] is true iff any of the following is true
         *      1) "x*" repeats 0 time and matches empty: f[i][j - 2]
         *      2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j]
         * '.' matches any single character
         */
        int m = s.size(), n = p.size();
        vector<vector<bool>> f(m + 1, vector<bool>(n + 1, false));
         
        f[0][0] = true;
        for (int i = 1; i <= m; i++)
            f[i][0] = false;
        // p[0.., j - 3, j - 2, j - 1] matches empty iff p[j - 1] is '*' and p[0..j - 3] matches empty
        for (int j = 1; j <= n; j++)
            f[0][j] = j > 1 && '*' == p[j - 1] && f[0][j - 2];
         
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++)
                if (p[j - 1] != '*')
                    f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]);
                else
                    // p[0] cannot be '*' so no need to check "j > 1" here
                    f[i][j] = f[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && f[i - 1][j];
         
        return f[m][n];
    }
};

　　

　　

　

类似题目：

[LeetCode] 44. Wildcard Matching 外卡匹配

All LeetCode Questions List 题目汇总

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