GitHub

leetcode_old/10.regular-expression-matching.cpp

1, If p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1]; 2, If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1]; 3, If p.charAt(j) == '*': here are two sub conditions ...
GitHub

# 010-regular-expression-matching

1. We check the first characters matched by s\[0\] == p\[0\] or p\[0\] == '.' 2. If the second character is \*, then there are two cases 1. We ignore the whether first character matches or not, we ...