LeetCode 794. Valid Tic-Tac-Toe State

原题链接在这里：https://leetcode.com/problems/valid-tic-tac-toe-state/

题目：

A Tic-Tac-Toe board is given as a string array board. Return True if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.

The board is a 3 x 3 array, and consists of characters " ", "X", and "O".  The " " character represents an empty square.

Here are the rules of Tic-Tac-Toe:

• Players take turns placing characters into empty squares (" ").
• The first player always places "X" characters, while the second player always places "O" characters.
• "X" and "O" characters are always placed into empty squares, never 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.

Example 1:
Input: board = ["O  ", "   ", "   "]
Output: false
Explanation: The first player always plays "X".

Example 2:
Input: board = ["XOX", " X ", "   "]
Output: false
Explanation: Players take turns making moves.

Example 3:
Input: board = ["XXX", "   ", "OOO"]
Output: false

Example 4:
Input: board = ["XOX", "O O", "XOX"]
Output: true

Note:

• board is a length-3 array of strings, where each string board[i] has length 3.
• Each board[i][j] is a character in the set {" ", "X", "O"}.

题解：

X player goes first. turn++.

O player goes next. turn--.

When checking the board, check row, col and diagonal, to see if there is n or -n. If yes, then x wins with n, o wins with -n.

When x wins turn must be 1. When o wins turn must be 0. If not, return false.

Check if turn is within 0 and 1.

Time Complexity: O(n ^ 2). n = board.length.

Space: O(n).

AC Java:

class Solution {
    public boolean validTicTacToe(String[] board) {
        int n = board.length;
        int turn = 0;
        boolean xWin = false;
        boolean oWin = false;
        int [] row = new int[n];
        int [] col = new int[n];
        int d = 0;
        int antiD = 0;
        
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(board[i].charAt(j) == 'X'){
                    turn++;
                    row[i]++;
                    col[j]++;
                    if(i == j){
                        d++;
                    }
                    
                    if(i + j == 2){
                        antiD++;
                    }
                }else if(board[i].charAt(j) == 'O'){
                    turn--;
                    row[i]--;
                    col[j]--;
                    if(i == j){
                        d--;
                    }
                    
                    if(i + j == 2){
                        antiD--;
                    }
                }
            }
        }
        
        for(int i = 0; i < n; i++){
            xWin |= row[i] == n;
            xWin |= col[i] == n;
            
            oWin |= row[i] == -n;
            oWin |= col[i] == -n;
        }
        
        xWin = xWin || d == n || antiD == n;
        oWin = oWin || d == -n || antiD == -n;
        
        if((xWin && turn != 1) || (oWin && turn != 0)){
            return false;
        }
        
        return turn == 0 || turn == 1;
    }
}

类似Design Tic-Tac-Toe.