Design Tic-Tac-Toe

题目地址：
https://leetcode.com/problems/design-tic-tac-toe/description/

题目：

Design a Tic-tac-toe game that is played between two players on a n x n grid.

You may assume the following rules:

1. A move is guaranteed to be valid and is placed on an empty block.
2. Once a winning condition is reached, no more moves is allowed.
3. 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?

解题思路：
这道题就是建立对于行和列以及两条对角线的count就好，先更新，然后只要达到size - 1就可以了。

代码：

class TicTacToe {

    class TwoPair{
        int firstCount;
        int secondCount;

        public TwoPair(){
            firstCount = 0;
            secondCount = 0;
        }
    }

    TwoPair[] rowCount;
    TwoPair[] colCount;
    TwoPair diaCount;            //    /    TwoPair reverseDiaCount;    //    \    int size;

    /** Initialize your data structure here. */    public TicTacToe(int n) {
        this.size = n;
        rowCount = new TwoPair[n];
        colCount = new TwoPair[n];
        for(int i = 0; i <= n - 1; i++){
            rowCount[i] = new TwoPair();
            colCount[i] = new TwoPair();
        }
        diaCount = new TwoPair();
        reverseDiaCount = new TwoPair();
    }

    /** Player {player} makes a move at ({row}, {col}).     @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. */    public int move(int row, int col, int player) {
        update(row, col, player);
        if(rowCount[row].firstCount == size || colCount[col].firstCount == size || diaCount.firstCount == size || reverseDiaCount.firstCount == size){
            return 1;
        }
        if(rowCount[row].secondCount == size || colCount[col].secondCount == size || diaCount.secondCount == size || reverseDiaCount.secondCount == size){
            return 2;
        }
        return 0;
    }

    private void update(int row, int col, int player) {
        if(player == 1){
            rowCount[row].firstCount += 1;
            colCount[col].firstCount += 1;
            if(row == col){
                diaCount.firstCount += 1;
            }
            if(row + col == size - 1){
                reverseDiaCount.firstCount += 1;
            }
        }
        else{
            rowCount[row].secondCount += 1;
            colCount[col].secondCount += 1;
            if(row == col){
                diaCount.secondCount += 1;
            }
            if(row + col == size - 1){
                reverseDiaCount.secondCount += 1;
            }
        }
    }

    public static void main(String[] args){
        TicTacToe ticTacToe = new TicTacToe(3);
        int rst = ticTacToe.move(0, 0, 1);
        System.out.println(rst);
    }
}

Follow up是： 不能一个玩家连续走两次。
那么可以每个TwoPair里面存一个set，储存之前出现过的index。有重复就让重新走。