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  • Tic Tac Toe

    Problem

    N*N matrix is given with input red or black.You can move horizontally, vertically or diagonally. If 3 consecutive samecolor found, that color will get 1 point. So if 4 red are vertically then pointis 2. Find the winner.

    Solution

    注意检查左上到右下, 右上到左下

     1 public static boolean ticTacToe(boolean[][] board) {
     2     if(board == null || board.length == 0)    return false;
     3     
     4     int m = board.length;
     5     int n = board[0].length;
     6     
     7     int red = 0, black = 0;
     8     for(int i=0; i<m; i++) {
     9         for(int j=0; j<n; j++) {
    10             if(board[i][j]) {
    11                 if(i<m-2 && board[i+1][j] && board[i+2][j])    red++;    //check vertical
    12                 if(j<n-2 && board[i][j+1] && board[i][j+2])    red++;    //check horizontal
    13                 if(i<m-2 && j<n-2 && board[i+1][j+1] && board[i+2][j+2])    red++;    //check diag
    14             }
    15             else {
    16                 if(i<m-2 && !board[i+1][j] && !board[i+2][j])    black++;    //check vertical
    17                 if(j<n-2 && !board[i][j+1] && !board[i][j+2])    black++;    //check horizontal
    18                 if(i<m-2 && j<n-2 && !board[i+1][j+1] && !board[i+2][j+2])    black++;    //check diag
    19             }
    20         }
    21     }
    22     
    23     for(int i=m-1; i>=0; i--) {
    24         for(int j=n-1; j>=0; j--) {
    25             if(board[i][j]) {
    26                 if(i>=2 && j>=2 && board[i-1][j-1] && board[i-2][j-2])    red++;    //check reverse diag
    27             }
    28             else {
    29                 if(i>=2 && j>=2 && !board[i-1][j-1] && !board[i-2][j-2])    black++;    //check reverse diag
    30             }
    31         }
    32     }
    33     
    34     return red > black;
    35 }
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  • 原文地址:https://www.cnblogs.com/superbo/p/4111925.html
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