leetcode 10. Regular Expression Matching

Given an input string (s) and a pattern §, 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 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 = “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

这个题用 DP 写，主要难点在于状态转移的确定以及初始值的确定。为了方便处理，统一在 s,p 前面加一个特殊字符，比如 #，令 $dp[i][j]$ 表示 $s[0,i]$ 和 $p[0,j]$ 匹配与否，则 $dp[0][0]=true$ （因为 0 处都是特殊字符 #） ，另外的初始化是：

$dp[0][j]=dp[0][j-2]\text{ and } p[j]==\verb|'*'|$

然后是状态转移：

1. s[i]==p[j]||p[j]=='.' ，则有 dp[i][j]=dp[i-1][j-1]
2. p[j]=='*' ，
   1. p[j-1]!=s[i]&&p[j-1]!='.' 则 '*' 作用必须是0：dp[i][j]=dp[i][j-2]
   2. dp[i][j]=dp[i][j-2]||dp[i][j-1]||dp[i-1][j] 分别表示 * 重复0、1、多次的情况

时间复杂度：$O(MN)$ （分别是主串与匹配串的长度）

class Solution {
public:
    bool dp[1000][1000];
    bool isMatch(string s, string p) {
        int slen=s.length(),plen=p.length();
        memset(dp,false,sizeof(dp));
        s="#"+s;p="#"+p;
        dp[0][0]=true;
        for(int j=1;j<=plen;j++)
            if(p[j]=='*'&&dp[0][j-2])
                dp[0][j]=true;
        for(int i=1;i<=slen;i++)
            for(int j=1;j<=plen;j++){
                if(s[i]==p[j]||p[j]=='.')dp[i][j]=dp[i-1][j-1];
                if(p[j]=='*'){
                    if(p[j-1]!=s[i]&&p[j-1]!='.')
                        dp[i][j]=dp[i][j-2];
                    else
                        dp[i][j]=dp[i][j-2]||dp[i][j-1]||dp[i-1][j];
                }
            }
        return dp[slen][plen];
    }
};