【MEDIUM】Design Tic-Tac-Toe

发布于: 2018-12-20 19:56
阅读: 82
评论: 0
喜欢: 0

问题

原题链接:https://leetcode.com/problems/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?

分析过程

  • 输入:下棋的位置
  • 输出:棋盘情况,胜负情况
  • 思路:两个选手,一个下棋时记录为 1,一个记录为 -1,判定胜负的方法为:每一行、每一列、对角线的值累加,为 n 或 -n,即游戏结束,否则游戏继续。

解决方法

class TicTacToe {
    
    var board = [[Int]]()
    
    var n: Int {
        return board.count
    }
    
    /** Initialize your data structure here. */
    init(_ n: Int) {
        for _ in 0..<n {
            board.append([Int].init(repeating: 0, count: n))
        }
    }
    
    /** 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. */
    func move(_ row: Int, _ col: Int, _ player: Int) -> Int {
        let playerToValue = [1 : -1, 2 : 1]
        board[row][col] = playerToValue[player]!
        return vaildBoard(row: row, col: col) ? player : 0
    }
    
    // 检查棋盘,返回 true 代表游戏结束
    internal func vaildBoard(row: Int, col: Int) -> Bool {
        // 行
        do {
            let sum = board[row].reduce(0) { $0 + $1 }
            if sum == n || sum == -n {
                return true
            }
        }
        // 列
        do {
            var sum = 0
            for field in 0..<n {
                sum += board[field][col]
            }
            if sum == n || sum == -n {
                return true
            }
        }
        // 扫描对角线
        var sum = 0
        for i in 0..<n {
            sum += board[i][i]
            if sum == n || sum == -n {
                return true
            }
        }
        sum = 0
        for i in 0..<n {
            sum += board[n - 1 - i][i]
            if sum == n || sum == -n {
                return true
            }
        }
        
        return false
    }
}

Thanks for reading.

All the best wishes for you! 💕