refactor 348

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_348.java

@@ -1,71 +1,10 @@
 package com.fishercoder.solutions;
 
-/**
- * 348. Design Tic-Tac-Toe
- *
- * Design a Tic-tac-toe game that is played between two players on a n x n grid.
-
- You may assume the following rules:
-
- A move is guaranteed to be valid and is placed on an empty block.
- Once a winning condition is reached, no more moves is allowed.
- A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.
-
- Example:
-
- Given n = 3, assume that player 1 is "X" and player 2 is "O" in the board.
-
- TicTacToe toe = new TicTacToe(3);
-
- toe.move(0, 0, 1); -> Returns 0 (no one wins)
- |X| | |
- | | | |    // Player 1 makes a move at (0, 0).
- | | | |
-
- toe.move(0, 2, 2); -> Returns 0 (no one wins)
- |X| |O|
- | | | |    // Player 2 makes a move at (0, 2).
- | | | |
-
- toe.move(2, 2, 1); -> Returns 0 (no one wins)
- |X| |O|
- | | | |    // Player 1 makes a move at (2, 2).
- | | |X|
-
- toe.move(1, 1, 2); -> Returns 0 (no one wins)
- |X| |O|
- | |O| |    // Player 2 makes a move at (1, 1).
- | | |X|
-
- toe.move(2, 0, 1); -> Returns 0 (no one wins)
- |X| |O|
- | |O| |    // Player 1 makes a move at (2, 0).
- |X| |X|
-
- toe.move(1, 0, 2); -> Returns 0 (no one wins)
- |X| |O|
- |O|O| |    // Player 2 makes a move at (1, 0).
- |X| |X|
-
- toe.move(2, 1, 1); -> Returns 1 (player 1 wins)
- |X| |O|
- |O|O| |    // Player 1 makes a move at (2, 1).
- |X|X|X|
-
- Follow up:
- Could you do better than O(n2) per move() operation?
-
- Hint:
-
- Could you trade extra space such that move() operation can be done in O(1)?
- You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.
-
- */
 public class _348 {
     public static class Solution1 {
         /**
          * credit: https://discuss.leetcode.com/topic/44548/java-o-1-solution-easy-to-understand
-         *
+         * <p>
          * Key: in order to win a TicTacToe, you must have the entire row or column, thus, we don't need
          * to keep track of the entire n^2 board. We only need to keep a count for each row and column.
          * If at any time, a row or column matches the size of the board, then that player has won.
@@ -97,8 +36,8 @@ public TicTacToe(int n) {
             /**
              * Player {player} makes a move at ({row}, {col}).
              *
-             * @param row The row of the board.
-             * @param col The column of the board.
+             * @param row    The row of the board.
+             * @param col    The column of the board.
              * @param player The player, can be either 1 or 2.
              * @return The current winning condition, can be either: 0: No one wins. 1: Player 1 wins. 2:
              * Player 2 wins.
@@ -118,9 +57,9 @@ public int move(int row, int col, int player) {
                 }
 
                 if (Math.abs(rows[row]) == size
-                    || Math.abs(cols[col]) == size
-                    || Math.abs(antidiagonal) == size
-                    || Math.abs(diagonal) == size) {
+                        || Math.abs(cols[col]) == size
+                        || Math.abs(antidiagonal) == size
+                        || Math.abs(diagonal) == size) {
                     return player;
                 }