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/* | |
This file contains the core code for dynamic programming | |
related to string | |
*/ | |
/* LC 10. Regular expression matching | |
dp (i,j): can first i char of s match first j char of p | |
case 1. p[j-1] != '*' and p[j-1] != '.' ==> function: dp[i-1][j-1] && s[i-1] == p[j-1] | |
case 2. p[j-1] == '.' ==> dp[i-1][j-1] | |
case 3. p[j-1] == '*' | |
3.a.p[j-2] != '.' and s[i-1] != p[j-2], (make p[j-2] disappear) ==> function: dp[i][j-2] | |
3.b. p[j-2] == '.': dp[i][j-2] (repeat 0 time) || dp[i][j-1] (repeat 1 time) || | |
dp[i-1][j] (repeat two or more) | |
*/ | |
bool isMatch(string s, string p){ | |
// initialize | |
// base case | |
// core code | |
for (int i = 1; i <= s.length(); ++i){ | |
for (int j = 1; j <= p.length(); ++j){ | |
if (p[j-1] == '.') { | |
dp[i][j] = dp[i-1][j-1]; | |
} else if (p[j-1] != '*'){ | |
dp[i][j] = dp[i-1][j-1] && s[i-1] == p[j-1]; | |
} else if (p[j-2] != '.' && s[i-1] != p[j-2]){ | |
dp[i][j] = dp[i][j-2]; | |
} else { | |
dp[i][j] = dp[i][j-2] || dp[i][j-1] || dp[i-1][j]; | |
} | |
} | |
} | |
return false; | |
} |
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