leetcode-1275-easy

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. The rules of Tic-Tac-Toe are:

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 three 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 a 2D integer array moves where moves[i] = [rowi, coli] indicates that the ith move will be played on grid[rowi][coli]. 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 (i.e., 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, they always play first.
Example 2:


Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
Output: "B"
Explanation: B wins.
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.
Constraints:

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

思路一：这题完全是体力活，完全分类有三种情况

• 分出胜负 -> 8 种
• 平局
• 未完全

对第一种分类的 8 种情况进行判断，如果分类不在第一种，只要判断棋盘是否有空余，就能区分另外的两种分类

public String tictactoe(int[][] moves) {
    int[][] arr = new int[3][3];

    boolean first = true;
    for (int[] move : moves) {
        arr[move[0]][move[1]] = first ? 1 : 2;
        first = !first;
    }

    for (int i = 0; i < 3; i++) {
        if (arr[i][0] != 0 && arr[i][0] == arr[i][1] && arr[i][1] == arr[i][2]) {
            return arr[i][0] == 1 ? "A" : "B";
        }
    }

    for (int i = 0; i < 3; i++) {
        if (arr[0][i] != 0 && arr[0][i] == arr[1][i] && arr[1][i] == arr[2][i]) {
            return arr[0][i] == 1 ? "A" : "B";
        }
    }

    if (arr[0][0] != 0 && arr[0][0] == arr[1][1] && arr[1][1] == arr[2][2]) {
        return arr[0][0] == 1 ? "A" : "B";
    }

    if (arr[2][0] != 0 && arr[2][0] == arr[1][1] && arr[1][1] == arr[0][2]) {
        return arr[2][0] == 1 ? "A" : "B";
    }

    for (int[] ints : arr) {
        for (int j = 0; j < arr[0].length; j++) {
            if (ints[j] == 0) {
                return "Pending";
            }
        }
    }

    return "Draw";
}