chore(Java): added longest common substring (#509)

Co-authored-by: Ming Tsai <37890026+ming-tsai@users.noreply.github.com>

algorithms/Java/README.md

@@ -76,7 +76,8 @@
 5. [Split String](strings/SplitString.java)
 6. [Tokenizer](strings/tokenizer.java)
 7. [Anagram](strings/anagram.java)
-8. [Boyer Moore Search](strings/Boyer_Moore.java)
+8. [Longest Common Substring](strings/Longest_common_substring.java)
+9. [Boyer Moore Search](strings/Boyer_Moore.java)
 
 ## Trees
 

algorithms/Java/strings/Longest_common_substring.java

@@ -0,0 +1,68 @@
+// Java implementation of
+// finding length of longest
+// Common substring using
+// Dynamic Programming
+class GFG {
+	/*
+	Returns length of longest common substring
+	of X[0..m-1] and Y[0..n-1]
+	*/
+	static int LCSubStr(char X[], char Y[], int m, int n)
+	{
+		// Create a table to store
+		// lengths of longest common
+		// suffixes of substrings.
+		// Note that LCSuff[i][j]
+		// contains length of longest
+		// common suffix of
+		// X[0..i-1] and Y[0..j-1].
+		// The first row and first
+		// column entries have no
+		// logical meaning, they are
+		// used only for simplicity of program
+		int LCStuff[][] = new int[m + 1][n + 1];
+	
+		// To store length of the longest
+		// common substring
+		int result = 0;
+
+		// Following steps build
+		// LCSuff[m+1][n+1] in bottom up fashion
+		for (int i = 0; i <= m; i++)
+		{
+			for (int j = 0; j <= n; j++)
+			{
+				if (i == 0 || j == 0)
+					LCStuff[i][j] = 0;
+				else if (X[i - 1] == Y[j - 1])
+				{
+					LCStuff[i][j] = LCStuff[i - 1][j - 1] + 1;
+					result = Integer.max(result, LCStuff[i][j]);
+				}
+				else
+					LCStuff[i][j] = 0;
+			}
+		}
+		return result;
+	}
+
+	// Driver Code
+	public static void main(String[] args)
+	{
+		// Test 1:
+		String X = "abcdabaabbacd";
+		String Y = "abbabaabcabcd";
+		int m = X.length();
+		int n = Y.length();
+		System.out.println(LCSubStr(X.toCharArray(), Y.toCharArray(), m, n));
+		
+		// Test 2:
+		String X = "abcdefgh";
+		String Y = "abaabcfg";
+		int m = X.length();
+		int n = Y.length();
+		System.out.println(LCSubStr(X.toCharArray(), Y.toCharArray(), m, n));
+	}
+}
+// Time Complexity -> O(m*n)
+// where m and n are the lengths of the strings X and Y.