[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:
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
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:
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:
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:
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:
# 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:
# 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:
# 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:
# 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:
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++:
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++:
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];
}
};
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