Assume the following rules are for the tic-tac-toe game on an n x n board between two players:
- A move is guaranteed to be valid and is placed on an empty block.
- Once a winning condition is reached, no more moves are allowed.
- A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.
Implement the TicTacToe class:
TicTacToe(int n)Initializes the object the size of the boardn.int move(int row, int col, int player)Indicates that the player with id player plays at the cell (row, col) of the board. The move is guaranteed to be a valid move.
Solution
按照正常思路，对每行，每列，对角线都 /(check/) 的话，每次都得 /(O(N)/). 但实际上我们只关心出现的次数是否为 /(n/) 即可，所以为每个 /(player/) 分别开个 /(vector/) 记录即可
点击查看代码
class TicTacToe {
private:
vector<vector<int>> R, C;
vector<int> diag, r_diag;
public:
TicTacToe(int n) {
R = vector<vector<int>> (n, vector<int>(3));
C = vector<vector<int>> (n, vector<int>(3));
diag = vector<int> (3);
r_diag = vector<int> (3);
}
int move(int row, int col, int player) {
int cnt;
R[row][player]++; C[col][player]++;
cnt = max(R[row][player], C[col][player]);
if(row==col)diag[player]++;
if(row+col==R.size()-1)r_diag[player]++;
cnt=max(cnt, max(diag[player], r_diag[player]));
if(cnt==R.size())return player;
else return 0;
}
};
/**
* Your TicTacToe object will be instantiated and called as such:
* TicTacToe* obj = new TicTacToe(n);
* int param_1 = obj->move(row,col,player);
*/
原创文章，作者：ItWorker，如若转载，请注明出处：https://blog.ytso.com/282538.html