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Tic Tac Toe

时间:2014-11-21 06:56:50      阅读:291      评论:0      收藏:0      [点我收藏+]

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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 }

 

Tic Tac Toe

标签:style   blog   io   ar   color   sp   for   strong   on   

原文地址:http://www.cnblogs.com/superbo/p/4111925.html

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