Create 10.Regular_expression_matching.java

Author: Amruta-22

Leetcode - Top Interview Questions/10.Regular_expression_matching.java

@@ -0,0 +1,40 @@
+class Solution {
+    public boolean isMatch(String s, String p) {
+        boolean[][] dp = new boolean[s.length() + 1][p.length()+1];
+        dp[0][0] = true;
+		
+		// initialize first column and row
+        for (int i = 1; i <= p.length(); i++) {
+            dp[0][i] = p.charAt(i - 1) == '*' && dp[0][i - 2];
+        }
+        
+        for (int i = 1; i <= s.length(); i++) {
+            dp[i][0] = false;
+        }
+        
+        for (int i = 1; i <= s.length(); i++) {
+            for (int j = 1; j <= p.length(); j++) {
+                if (p.charAt(j - 1) == '*') {
+					// we use * to match current character in string i
+					// aaa  -> aa   dp[i][j] = dp[i-1][j]
+					// a*   -> a*
+                    if (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') {
+                        dp[i][j] = dp[i-1][j];
+                    }
+					// ?* match 0 cahracters
+					// aa
+					// aab*
+                    dp[i][j] = dp[i][j - 2] || dp[i][j];
+                } else {
+				// ease case
+                    if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.') {
+                        dp[i][j] = dp[i - 1][j - 1];
+                    }
+                }
+            }
+        }
+        return dp[s.length()][p.length()];
+    }
+    
+    
+}