地址：https://leetcode.com/problems/regular-expression-matching/
题目：
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 = “cab”
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 = “misisP*.”
Output: false
理解：
字符串的正则匹配，*可以匹配0个或多个字符，.可以匹配一个任意字符
思路是从头开始匹配，先看s[i]和p[j]是不是相同，或者p[j]=='.'，如果是，当前位置是匹配的。
‘*’如果有，必然出现在j+1的位置。
如果p[j+1]=='*'，需要判断两种情况。
s[i]开始的字符串和p[j+2]开始的字符串是否匹配，即*匹配了0个字符；
或者s[i]和p[j]是否相同，且s[i+1]和p[j]是否相同，即*匹配了一个字符。
否则，判断s[i]和p[j]是否相同，且s[i+1]和p[j+1]是否相同。
实现1：
最简单的递归实现，由于不停的复制string，效率很低
class Solution {
public:
bool isMatch(string s,string p) {
if (p.empty()) return s.empty();
bool first_match = (!s.empty() && (p.at(0) == s.at(0) || p.at(0) == '.'));
if (p.length() >= 2 && p.at(1) == '*') {
return isMatch(s,p.substr(2)) || (first_match && isMatch(s.substr(1),p));
}
else
return first_match&& isMatch(s.substr(1),p.substr(1));
}
};
实现2：
使用下标，代替了字符串复制
enum class Match {
FALSE,TRUE,UNKNowN
};
class Solution {
vector
public:
bool isMatch(string s,string p) {
result = vector
return dp(0,s,p);
}
bool dp(int i,int j,const string& s,const string& p) {
if (result[i][j] != Match::UNKNowN)
return result[i][j] == Match::TRUE;
bool res;
if (j == p.length())
res = (i == s.length());
else {
bool first_match = i < s.length() && (s.at(i) == p.at(j) || p.at(j) == '.');
if (j + 1 < p.length()&&p.at(j+1)=='*') {
res = dp(i,j + 2,p) || (first_match&&dp(i + 1,j,p));
}
else {
res = first_match&&dp(i + 1,j + 1,p);
}
}
result[i][j] = res ? Match::TRUE : Match::FALSE;
return res;
}
};
实现3：
上面的实现2，还是使用递归的方式，然而前面的位置是与后面的位置有关的，因此可以从后向前，记录后面的判断结果，避免递归调用
class Solution {
vector
public:
bool isMatch(string s,string p) {
dp = vector
dp[s.length()][p.length()] = true;
for (int i = s.length(); i >= 0; --i)
for (int j = p.length() - 1; j >= 0; --j) {
bool first_match = i < s.length() && (s.at(i) == p.at(j) || p.at(j) == '.');
if (j + 1 < p.length() && p.at(j + 1) == '*') {
dp[i][j] = dp[i][j + 2] || (first_match &&dp[i + 1][j]);
}
else {
dp[i][j] = (first_match && dp[i + 1][j + 1]);
}
}
return dp[0][0];
}
};
