Leetcode_1275_Find Winner on a Tic Tac Toe Game

[lihe]

Find Winner on a Tic Tac Toe Game

Tic-tac-toe is played by two players A and B on a 3 x 3 grid.

Here are the rules of Tic-Tac-Toe:

• Players take turns placing characters into empty squares (" ").
• The first player A always places "X" characters, while the second player B always places "O" characters.
• "X" and "O" characters are always placed into empty squares, never on filled ones.
• The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
• The game also ends if all squares are non-empty.
• No more moves can be played if the game is over.

Given an array moves where each element is another array of size 2 corresponding to the row and column of the grid where they mark their respective character in the order in which A and B play.

Return the winner of the game if it exists (A or B), in case the game ends in a draw return "Draw", if there are still movements to play return "Pending".

You can assume that moves is valid (It follows the rules of Tic-Tac-Toe), the grid is initially empty and A will play first.

Example 1:

Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
Output: "A"
Explanation: "A" wins, he always plays first.
"X  "    "X  "    "X  "    "X  "    "X  "
"   " -> "   " -> " X " -> " X " -> " X "
"   "    "O  "    "O  "    "OO "    "OOX"

Example 2:

Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
Output: "B"
Explanation: "B" wins.
"X  "    "X  "    "XX "    "XXO"    "XXO"    "XXO"
"   " -> " O " -> " O " -> " O " -> "XO " -> "XO " 
"   "    "   "    "   "    "   "    "   "    "O  "

Example 3:

Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
Output: "Draw"
Explanation: The game ends in a draw since there are no moves to make.
"XXO"
"OOX"
"XOX"

Example 4:

Input: moves = [[0,0],[1,1]]
Output: "Pending"
Explanation: The game has not finished yet.
"X "
" O "
" "

Constraints:

• 1 <= moves.length <= 9
• moves[i].length == 2
• 0 <= moves[i][j] <= 2
• There are no repeated elements on moves.
• moves follow the rules of tic tac toe.

算法思路：

1. 一共只有九个格子，并且赢面是固定的，可以使用一个9位二进制代表最后棋盘的结果，规定：井字形棋盘的[i,j]代表二进制数的第3*i + j位，没走一步棋等价于与对应的位进行异或运算。
2. 将一方的数字与能获胜的数字k进行与运算，如果结果为k，则此方获胜，若双方都未获胜则：
   • 若总步数为9步，则平局；
   • 否则，未完成；
3. 赢面数字只有8种分别是7, 56, 448, 73, 146, 292, 273, 84

public String tictactoe(int[][] moves) {
    int a = 0, b = 0, len = moves.length;

    int[] ac =  {7, 56, 448, 73, 146, 292, 273, 84};
    for(int i = 0; i < len; i++){
        if((i & 1) == 1){
            b ^= 1 << (3 * moves[i][0] + moves[i][1]);
        }else{
            a ^= 1 << (3 * moves[i][0] + moves[i][1]);
        }
    }

    for(int i : ac){
        if((a & i) == i){
            return "A";
        }
        if((b & i) == i){
            return "B";
        }
    }
    return len == 9 ? "Draw" : "Pending";
}