348. Design Tic-Tac-Toe

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:

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?

Hint:

Could you trade extra space such that move() operation can be done in O(1)?
You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.

解答：
一开始其实就想到了hint, 作出了下面的解法：

public class TicTacToe {
int[][] grid;
int n;
/ Initialize your data structure here. */
public TicTacToe(int n) {
grid = new int[n][n];
this.n = n;
}

public boolean check(int row, int col, int len) {    boolean hori = true, verti = true, diag1 = true, diag2 = true;    //check horizontal    for (int i = 0; i 

这个解法冗余在check行个列的时候，每一次都要再扫一遍这一行这一列，所以如果只有两个player，可以把这两个player记作1, －1。当有一行完全只有这两个player中的其中一个人时，sum的绝对值应该等于这个数列的长度，这样就不需要每次再扫一遍数组。代码如下：

public class TicTacToe {
int[] rows, cols;
int diagonal, antiDiagonal;
int len;
/ Initialize your data structure here. */
public TicTacToe(int n) {
rows = new int[n];
cols = new int[n];
this.len = 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. */public int move(int row, int col, int player) {    //Take player 1 and 2 as value of 1 and -1;    //Every time we only do adding, dont need to re-scan the whole line    int toAdd = player == 1 ? 1 : -1;    rows[row] += toAdd;    cols[col] += toAdd;    if (row == col) diagonal += toAdd;    if (row == len - 1 - col) antiDiagonal += toAdd;    if (Math.abs(rows[row]) == len || Math.abs(cols[col]) == len || Math.abs(diagonal) == len || Math.abs(antiDiagonal) == len) {        return player;    }    return 0;}

}

关键字：java, leetcode

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