289. Life of Game

chuyao 發(fā)布于2019-08-14 16:31 / 2719人閱讀

摘要：?jiǎn)栴}解答我的解法是需要最多的空間的。如果要做到那么我看到里一個(gè)非常的做法是用一個(gè)的數(shù)表示改變前和改變后的狀態(tài)。

問題：
According to the Wikipedia"s article: "The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970."

Given a board with m by n cells, each cell has an initial state live (1) or dead (0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):

Any live cell with fewer than two live neighbors dies, as if caused by under-population.
Any live cell with two or three live neighbors lives on to the next generation.
Any live cell with more than three live neighbors dies, as if by over-population..
Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.
Write a function to compute the next state (after one update) of the board given its current state.

Follow up:
Could you solve it in-place? Remember that the board needs to be updated at the same time: You cannot update some cells first and then use their updated values to update other cells.
In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches the border of the array. How would you address these problems?

解答：
我的解法是需要最多o(mn)的空間的。

class Point {
    int x, y;
    public Point(int x, int y) {
        this.x = x;
        this.y = y;
    }
}

public void checkAlive(int[][] board, int val, int i, int j, List list) {
    int count = 0;
    int[][] dir = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
    for (int k = 0; k < dir.length; k++) {
        int x = i + dir[k][0], y = j + dir[k][1];
        if (x < 0 || x > board.length - 1 || y < 0 || y > board[0].length - 1) continue;
        count += board[x][y];
    }
    if (val == 1 && (count < 2 || count > 3)) {
        list.add(new Point(i, j));
    }
    if (val == 0 && count == 3) {
        list.add(new Point(i, j));
    }
}

public void gameOfLife(int[][] board) {
    if (board == null || board.length == 0 || board[0].length == 0) return;
    int m = board.length, n = board[0].length;
    List list = new ArrayList();
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            checkAlive(board, board[i][j], i, j, list);
        }
    }
    for (int i = 0; i < list.size(); i++) {
        Point p = list.get(i);
        board[p.x][p.y] = board[p.x][p.y] == 1 ? 0 : 1;
    }
}

如果要做到in space, 那么我看到discussion里一個(gè)非常tricky的做法是用一個(gè)2-bit的數(shù)表示改變前和改變后的狀態(tài)。改變前有兩個(gè)狀態(tài)：00, 01. 在滿足一定的條件后，改變前一個(gè)bit的狀態(tài)也有兩種：10, 11.所以我們只要記錄改變的狀態(tài)，然后將實(shí)際的數(shù)向后移動(dòng)一位就可以得到當(dāng)然的一狀態(tài)。

public int helper(int[][] board, int i, int j) {
    int count = 0;
    int[][] dir = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
    for (int k = 0; k < dir.length; k++) {
        int x = i + dir[k][0], y = j + dir[k][1];
        if (x < 0 || x > board.length - 1 || y < 0 || y > board[0].length - 1) continue;
        count += board[x][y] & 1;
    }
    return count;
}
public void gameOfLife(int[][] board) {
    if (board == null || board.length == 0 || board[0].length == 0) return;
    int m = board.length, n = board[0].length;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            int lives = helper(board, i, j);
            if (board[i][j] == 1 && (lives == 2 || lives == 3)) {
                board[i][j] = 3;
            }
            if (board[i][j] == 0 && lives == 3) {
                board[i][j] = 2;
            }
        }
    }
    
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            board[i][j] >>= 1;
        }
    }
}