LeetCode #10: Regular Expression Matching

Problem Statement

(Source) 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).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

Analysis

dp[i][j]

: the matching result of

s[:i]

and

p[:j]

.

Time Complexity: O(mn)

Space Complexity: O(mn)

where m and n are the length of s and p respectively.

Tags:

Dynamic Programming

.

Solution

class Solution(object):
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""

m, n = len(s), len(p)
dp = [[False for j in xrange(n + 1)] for i in xrange(m + 1)]
dp[0][0] = True

# Assume valid RE doesn't have '*' as the first character.
for j in xrange(1, n + 1):
if p[j - 1] == '*':
dp[0][j] = dp[0][j - 1]
if j > 1:
dp[0][j] = (dp[0][j] or dp[0][j - 2])

# If string is empty, then try to match p with empty string.
if m == 0:
return dp[0]

# Fill in the dp matrix.
for i in xrange(0, m + 1):
for j in xrange(1, n + 1):
if p[j - 1] == '.' or p[j - 1] == s[i - 1]:
dp[i][j] = dp[i - 1][j - 1]
elif p[j - 1] == '*':
if dp[i][j - 2]:
dp[i][j] = True
elif dp[i - 1][j] and (s[i - 1] == p[j - 2] or p[j - 2] == '.'):
dp[i][j] = True

return dp[m]

References

(1) http://bangbingsyb.blogspot.com.au/2014/11/leetcode-regular-expression-matching.html

(2) http://articles.leetcode.com/regular-expression-matching