• 348. Design Tic-Tac-Toe


    package LeetCode_348
    
    /**
     * 348. Design Tic-Tac-Toe
     * (Lock by leetcode)
     * https://www.lintcode.com/problem/design-tic-tac-toe/description
     * 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.
     * */
    class Solution {
        var matrix: Array<IntArray>? = null
        var rows: IntArray? = null
        var cols: IntArray? = null
        var dia1 = 0
        var dia2 = 0
        var n = 0
    
        fun TicTacToe(n: Int) {
            this.n = n
            matrix = Array(n, { IntArray(n) })
            rows = IntArray(n)
            cols = IntArray(n)
        }
    
        fun move2(row: Int, col: Int, player: Int): Int {
            val value = if (player == 1) 1 else -1
            rows!![row] += value
            cols!![col] += value
            //diagonal from top left to right bottom
            if (row == col) {
                dia1++
            }
            //diagonal from top right to left bottom
            if (col == n - row - 1) {
                dia2++
            }
    
            if (Math.abs(rows!![row]) == n ||
                Math.abs(cols!![col]) == n ||
                Math.abs(dia1) == n ||
                Math.abs(dia2) == n) {
                return player
            }
    
            return 0
        }
    
        /** 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. */
        fun move(row: Int, col: Int, player: Int): Int {
            /*
            * solution 1:Time complexity: O(n*n), Space complexity:O(n*n)
            * */
            val localMatrix = matrix
            localMatrix!![row][col] = player
            //check row
            var win = true
            for (i in localMatrix.indices) {
                if (localMatrix[row][i] != player) {
                    win = false
                    break
                }
            }
            if (win) {
                return player
            }
    
            //check columns
            win = true
            for (i in localMatrix.indices) {
                if (localMatrix[i][col] != player) {
                    win = false
                    break
                }
            }
            if (win) {
                return player
            }
    
            //check back diagonal
            win = true
            for (i in localMatrix.indices) {
                if (localMatrix[i][i] != player) {
                    win = false
                    break
                }
            }
            if (win) {
                return player
            }
    
            //check forward diagonal
            win = true
            for (i in localMatrix.indices) {
                if (localMatrix[i][localMatrix.size - i - 1] != player) {
                    win = false
                    break
                }
            }
            if (win) {
                return player
            }
    
            return 0
        }
    }
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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/13182522.html
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