题目来源：https://leetcode.com/problems/regular-expression-matching/
问题描述
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:
- 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
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题意
正则表达式匹配。其中’.’表示匹配一个任意字符，’x*’连起来表示x的任意次（包括0次）重复（x可以是a-z的字母，也可以是’.’，如果x是’.’则’.*’可以匹配任意字符串<包括空字符串>）。
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思路
难点在于对于’x*’的处理。
读入’x’时，判断紧跟在后面的字符是不是’*’，如果是，则转入’x*’的判断。对于’x*’，递归调用isMatch判断可能出现的各种匹配。
特别注意当原字符串s已经匹配完而模板字符串p仍有余的情况，这是如果p的余量都是’x*’的组合，则仍可能匹配成功。这种情况需要另外加以判断，同样可以用递归实现，也可以用循环实现。
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代码
class Solution {
public boolean isMatch(String s, String p) {
char sch = 0, pch = 0, pre = 0, next = 0;
int n = s.length(), m = p.length(), i = 0, j = 0, k = 0;
if (s.isEmpty() && p.isEmpty())
{
return true;
}
else if (!s.isEmpty() && p.isEmpty())
{
return false;
}
else if (s.isEmpty() && !p.isEmpty())
{
if (p.length() >= 2 && p.charAt(1) == '*')
{
return isMatch(s, p.substring(2));
}
else
{
return false;
}
}
while (i < n && j < m)
{
sch = s.charAt(i);
pch = p.charAt(j);
if (pch >= 'a' && pch <= 'z')
{
if (j < m-1 && p.charAt(j+1) == '*')
{
pre = pch;
j++;
}
else
{
if (sch != pch)
{
return false;
}
else
{
pre = pch;
i++;
j++;
}
}
}
else if (pch == '.')
{
if (j < m-1 && p.charAt(j+1) == '*')
{
pre = pch;
j++;
}
else
{
pre = pch;
i++;
j++;
}
}
else
{
if (pre >= 'a' && pre <= 'z')
{
if (isMatch(s.substring(i), p.substring(j+1)))
{
return true;
}
k = i;
while (k < n && pre == s.charAt(k))
{
if (isMatch(s.substring(k+1), p.substring(j+1)))
{
return true;
}
k++;
}
return false;
}
else if (pre == '*')
{
return false;
}
else if (pre == 0)
{
return false;
}
else
{
if (isMatch(s.substring(i), p.substring(j+1)))
{
return true;
}
k = i;
while (k < n)
{
if (isMatch(s.substring(k+1), p.substring(j+1)))
{
return true;
}
k++;
}
return false;
}
}
}
if (i < n && j == m)
{
return false;
}
else if (i == n && j == m)
{
return true;
}
else if (i == n && j < m)
{
return isMatch("", p.substring(j));
}
return false;
}
}