Face questions 19. The regular expression matching --leecode

topic

Implement comprises a function to match '' and ' n' of expression. Mode character '.' Represents any one character ' ' is represented in front of the character may be any number of times (including 0 time). In this problem, the match is a whole pattern matches all characters of the string. For example, the string "aaa" mode and "aa" and "ab & AC A" match, but "aa.a" and "ab * a" do not match.
Topic Link

Thinking

Dynamic programming:
DP [i] [j] denotes the first j characters before i-th character of s and p are matched.
Note pit:
1. The beginning of the initialization, in particular dp [0] [m] values.
2. To open up an array bigger.
3. Match 0 really is a giant pit!

Code

class Solution {
public:
    bool isMatch(string s, string p) {
        int l1 = s.size(),l2 = p.size();
        vector<vector<bool> > dp(l1 + 1,vector<bool>(l2 + 1,false));
        dp[0][0] = true;
        for(int j = 1;j <= l2;j++){
            if(p[j - 1] == '*' && j - 2 >= 0) dp[0][j] = dp[0][j - 2];
        }
        for(int i = 1;i <= l1;i++){
            for(int j = 1;j <= l2;j++){
                if(p[j - 1] == s[i - 1] || p[j - 1] == '.'){
                    dp[i][j] = dp[i - 1][j - 1];
                }else if(p[j - 1] == '*'){
                    if(p[j - 2] == s[i - 1] || p[j - 2] == '.')
                        dp[i][j] = (dp[i - 1][j] || dp[i][j - 2]);
                    else
                        dp[i][j] = dp[i][j - 2];
                }
            }
        }
        return dp[l1][l2];
    }
};