refactor 419

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_419.java

@@ -1,10 +1,17 @@
 package com.fishercoder.solutions;
 
-/**Given an 2D board, count how many battleships are in it. The battleships are represented with 'X's, empty slots are represented with '.'s. You may assume the following rules:
+/**
+ * 419. Battleships in a Board
+ *
+ * Given an 2D board, count how many battleships are in it.
+ * The battleships are represented with 'X's, empty slots are represented with '.'s.
+ * You may assume the following rules:
 
  You receive a valid board, made of only battleships or empty slots.
- Battleships can only be placed horizontally or vertically. In other words, they can only be made of the shape 1xN (1 row, N columns) or Nx1 (N rows, 1 column), where N can be of any size.
+ Battleships can only be placed horizontally or vertically.
+ In other words, they can only be made of the shape 1xN (1 row, N columns) or Nx1 (N rows, 1 column), where N can be of any size.
  At least one horizontal or vertical cell separates between two battleships - there are no adjacent battleships.
+
  Example:
 
  X..X
@@ -27,86 +34,82 @@
  */
 public class _419 {
 
-    /**
-     * credit: https://discuss.leetcode.com/topic/62970/simple-java-solution,
-     * basically, it only counts the top-left one while ignoring all other parts of one battleship,
-     * using the top-left one as a representative for one battle.
-     * This is achieved by counting cells that don't have 'X' to the left and above them.
-     */
-    public int countBattleships_no_modify_original_input(char[][] board) {
-        if (board == null || board.length == 0) {
-            return 0;
-        }
-        int count = 0;
-        int m = board.length;
-        int n = board[0].length;
-        for (int i = 0; i < m; i++) {
-            for (int j = 0; j < n; j++) {
-                if (board[i][j] == '.') {
-                    continue;//if it can pass this line, then board[i][j] must be 'X'
-                }
-                if (j > 0 && board[i][j - 1] == 'X') {
-                    continue;//then we check if its left is 'X'
-                }
-                if (i > 0 && board[i - 1][j] == 'X') {
-                    continue;//also check if its top is 'X'
+    public static class Solution1 {
+        /**
+         * credit: https://discuss.leetcode.com/topic/62970/simple-java-solution,
+         * <p>
+         * This solution does NOT modify original input.
+         * Basically, it only counts the top-left one while ignoring all other parts of one battleship,
+         * using the top-left one as a representative for one battle.
+         * This is achieved by counting cells that don't have 'X' to the left and above them.
+         */
+        public int countBattleships(char[][] board) {
+            if (board == null || board.length == 0) {
+                return 0;
+            }
+            int count = 0;
+            int m = board.length;
+            int n = board[0].length;
+            for (int i = 0; i < m; i++) {
+                for (int j = 0; j < n; j++) {
+                    if (board[i][j] == '.') {
+                        continue;//if it can pass this line, then board[i][j] must be 'X'
+                    }
+                    if (j > 0 && board[i][j - 1] == 'X') {
+                        continue;//then we check if its left is 'X'
+                    }
+                    if (i > 0 && board[i - 1][j] == 'X') {
+                        continue;//also check if its top is 'X'
+                    }
+                    count++;
                 }
-                count++;
             }
+            return count;
         }
-        return count;
     }
 
-    /**
-     * My original solution, actually modified the input. I just undo it at the end.
-     */
-    public int countBattleships(char[][] board) {
-        if (board == null || board.length == 0) {
-            return 0;
-        }
-        int result = 0;
-        int m = board.length;
-        int n = board[0].length;
-        for (int i = 0; i < m; i++) {
-            for (int j = 0; j < n; j++) {
-                if (board[i][j] == 'X') {
-                    result++;
-                    dfs(board, i, j, m, n);
+    public static class Solution2 {
+        /**
+         * My original solution, actually modified the input. I just undo it at the end.
+         */
+        public int countBattleships(char[][] board) {
+            if (board == null || board.length == 0) {
+                return 0;
+            }
+            int result = 0;
+            int m = board.length;
+            int n = board[0].length;
+            for (int i = 0; i < m; i++) {
+                for (int j = 0; j < n; j++) {
+                    if (board[i][j] == 'X') {
+                        result++;
+                        dfs(board, i, j, m, n);
+                    }
                 }
             }
-        }
 
-        for (int i = 0; i < m; i++) {
-            for (int j = 0; j < n; j++) {
-                if (board[i][j] == '#') {
-                    board[i][j] = 'X';
+            for (int i = 0; i < m; i++) {
+                for (int j = 0; j < n; j++) {
+                    if (board[i][j] == '#') {
+                        board[i][j] = 'X';
+                    }
                 }
             }
+            return result;
         }
-        return result;
-    }
 
-    private void dfs(char[][] board, int x, int y, int m, int n) {
-        if (x < 0 || x >= m || y < 0 || y >= n || board[x][y] != 'X') {
-            return;
-        }
-        if (board[x][y] == 'X') {
-            board[x][y] = '#';
+        private void dfs(char[][] board, int x, int y, int m, int n) {
+            if (x < 0 || x >= m || y < 0 || y >= n || board[x][y] != 'X') {
+                return;
+            }
+            if (board[x][y] == 'X') {
+                board[x][y] = '#';
+            }
+            dfs(board, x + 1, y, m, n);
+            dfs(board, x, y + 1, m, n);
+            dfs(board, x - 1, y, m, n);
+            dfs(board, x, y - 1, m, n);
         }
-        dfs(board, x + 1, y, m, n);
-        dfs(board, x, y + 1, m, n);
-        dfs(board, x - 1, y, m, n);
-        dfs(board, x, y - 1, m, n);
     }
 
-    public static void main(String... strings) {
-        char[][] board = new char[][]{
-                {'X', '.', '.', 'X'},
-                {'.', '.', '.', 'X'},
-                {'.', '.', '.', 'X'},
-        };
-
-        _419 test = new _419();
-        System.out.println(test.countBattleships(board));
-    }
-}
+}