LeetCode解题 10：Regular Expression Matching（动态规划解法）
Problem 10: Regular Expression Matching [Hard]
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 preceding 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
来源：LeetCode
解题思路
使用动态规划思想。维护数组match[n][m]，match[i][j]代表字符串s的前i个和模式p的前j个是否匹配。二重循环遍历s和p，match[i][j]根据p[j]和s[i]的关系可以由match[i-1][j-1]等前项转换而来，具体转换公式为：
-
当
p
[
j
]
=
s
[
i
]
p[j] = s[i]
p[j]=s[i] 或者
p
[
j
]
=
p[j] =
p[j]=
'.'时，有
m
a
t
c
h
[
i
]
[
j
]
=
m
a
t
c
h
[
i
−
1
]
[
j
−
1
]
match[i][j] = match[i-1][j-1]
match[i][j]=match[i−1][j−1]
-
当
p
[
j
]
=
p[j] =
p[j]=
'*'时，情况比较复杂。需要考虑*的前一个字符，即p
[
j
−
1
]
p[j-1]
p[j−1]。
2.1 若
p
[
j
−
1
]
≠
s
[
i
]
p[j-1] \neq s[i]
p[j−1]=s[i] 且
p
[
j
−
1
]
≠
p[j-1] \neq
p[j−1]=
'.'，则*只能匹配零个，相当于去掉p
[
j
−
1
]
p[j-1]
p[j−1]和
p
[
j
]
p[j]
p[j]，转换公式为：
m
a
t
c
h
[
i
]
[
j
]
=
m
a
t
c
h
[
i
]
[
j
−
2
]
match[i][j] = match[i][j-2]
match[i][j]=match[i][j−2]
2.2 若p
[
j
−
1
]
=
s
[
i
]
p[j-1] = s[i]
p[j−1]=s[i] 或者
p
[
j
−
1
]
=
p[j-1] =
p[j−1]=
'.'，那么*有可能匹配零个或多个。假设当前s为###a，p为###a*，则match[i][j]有可能从不同的前项转移而来：i.
*匹配多个a的情况，例如：czha|a和czha|*。此时s[i]上的a只是a*的延续，转移公式为：
m
a
t
c
h
[
i
]
[
j
]
=
m
a
t
c
h
[
i
−
1
]
[
j
]
match[i][j] = match[i-1][j]
match[i][j]=match[i−1][j]
注意：除了上述例子中*匹配了两个a的情况，转移公式同样适用于匹配一个a的情况。（s =czh|a，p =czha|*，match[i-1][j]代表czh和czha*匹配，仍然可以完全匹配。）ii.
*匹配零个a的情况，例如：czh|a和czhaa|*。此时a*是多余的，可以去掉，转移公式为：
m
a
t
c
h
[
i
]
[
j
]
=
m
a
t
c
h
[
i
]
[
j
−
2
]
match[i][j] = match[i][j-2]
match[i][j]=match[i][j−2]
情况i和情况ii中，只要任意一个能匹配上，则m
a
t
c
h
[
i
]
[
j
]
=
t
r
u
e
match[i][j] = true
match[i][j]=true。
-
当
p
[
j
]
p[j]
p[j]不属于上述任何情况，即
p
[
j
]
p[j]
p[j]不为特殊符号且
p
[
j
]
≠
s
[
i
]
p[j] \neq s[i]
p[j]=s[i]时，
m
a
t
c
h
[
i
]
[
j
]
=
f
a
l
s
e
match[i][j] = false
match[i][j]=false。
思路参考LeetCode题解
时间复杂度为
O
(
n
m
)
O(nm)
O(nm)，空间复杂度为
O
(
n
m
)
O(nm)
O(nm)。
运行结果：
要点：动态规划
Solution (Java)
class Solution {
public boolean isMatch(String s, String p) {
int n = s.length() + 1;
int m = p.length() + 1;
if(n == 1 && m == 1) return true;
if(n > 1 && m == 1) return false;
boolean[][] match = new boolean[n][m];
// initialize
match[0][0] = true;
match[0][1] = false;
for(int mj = 2; mj < m; mj++){
int j = mj - 1;
if(p.charAt(j) == '*') match[0][mj] = match[0][mj-2];
else match[0][mj] = false;
}
for(int mi = 1; mi < n; mi++){
match[mi][0] = false;
}
// dp
for(int mi = 1; mi < n; mi++){
for(int mj = 1; mj < m; mj++){
int i = mi - 1;
int j = mj - 1;
if(p.charAt(j) == s.charAt(i) || p.charAt(j) == '.'){
match[mi][mj] = match[mi-1][mj-1];
}
else if(p.charAt(j) == '*'){
if(p.charAt(j-1) == s.charAt(i) || p.charAt(j-1) == '.'){
match[mi][mj] = match[mi-1][mj] || match[mi][mj-2];
}
else{
match[mi][mj] = match[mi][mj-2];
}
}
else{
match[mi][mj] = false;
}
}
}
return match[n-1][m-1];
}
}