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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 vertical12 if(j<n-2 && board[i][j+1] && board[i][j+2]) red++; //check horizontal13 if(i<m-2 && j<n-2 && board[i+1][j+1] && board[i+2][j+2]) red++; //check diag14 }15 else {16 if(i<m-2 && !board[i+1][j] && !board[i+2][j]) black++; //check vertical17 if(j<n-2 && !board[i][j+1] && !board[i][j+2]) black++; //check horizontal18 if(i<m-2 && j<n-2 && !board[i+1][j+1] && !board[i+2][j+2]) black++; //check diag19 }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 diag27 }28 else {29 if(i>=2 && j>=2 && !board[i-1][j-1] && !board[i-2][j-2]) black++; //check reverse diag30 }31 }32 }33 34 return red > black;35 }
Tic Tac Toe
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