leetcode289- Game of Life- medium

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):

 

1. Any live cell with fewer than two live neighbors dies, as if caused by under-population.
2. Any live cell with two or three live neighbors lives on to the next generation.
3. Any live cell with more than three live neighbors dies, as if by over-population..
4. 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: 

1. 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.
2. 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?

 

 

1.用额外空间的。先根据每个live向四面八方做贡献，给周围格子live计数++。从而先得到整个棋盘每个点的四周live计数矩阵。之后根据该矩阵和条件更新状态。

2.利用2位记录状态。00,01,10,11，低位表示当前状态，高位表示下轮状态。这样计数的时候只看board[i][j] & 1即可。就算把1改2（下轮要变0）也不会影响后续计数。

 

 

实现1：

class Solution {
    public void gameOfLife(int[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) {
            return;
        }
        int[][] count = new int[board.length][board[0].length];
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                count(board, i, j, count);
            }
        }
        
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                int cnt = count[i][j];
                if (board[i][j] == 0 && cnt == 3) {
                    board[i][j] = 1;
                } else if (board[i][j] == 1 && (cnt < 2 || cnt > 3)) {
                    board[i][j] = 0;
                }
            }
        }
    }

    private void count(int[][] board, int x, int y, int[][] count) {
        if (board[x][y] == 0) {
            return;
        }
        int[] dx = {0, 1, 0, -1, 1, 1, -1, -1};
        int[] dy = {1, 0, -1, 0, 1, -1, 1, -1};
        for (int i = 0; i < 8; i++) {
            int newX = x + dx[i];
            int newY = y + dy[i];
            if (isInBound(board, newX, newY)) {
                count[newX][newY]++;
            }
        }
    }

    private boolean isInBound(int[][] board, int x, int y) {
        return x >= 0 && x < board.length && y >= 0 && y < board[0].length;
    }
}

 

2.实现2

class Solution {
    public void gameOfLife(int[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) {
            return;
        }
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                int cnt = count(board, i, j);
                if (board[i][j] == 0 && cnt == 3) {
                    board[i][j] = 2;
                } else if (board[i][j] == 1 && (cnt < 2 || cnt > 3)) {
                    board[i][j] = 5;
                } else if (board[i][j] == 1) {
                    board[i][j] = 3;
                }
            }
        }

        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                board[i][j] = ((board[i][j] >> 1) & 1);
            }
        }
    }

    private int count(int[][] board, int x, int y) {
        int[] dx = {0, 1, 0, -1, 1, 1, -1, -1};
        int[] dy = {1, 0, -1, 0, 1, -1, 1, -1};
        int cnt = 0;
        for (int i = 0; i < 8; i++) {
            int newX = x + dx[i];
            int newY = y + dy[i];
            if (isInBound(board, newX, newY) && (board[newX][newY] & 1) == 1) {
                cnt++;
            }
        }
        return cnt;
    }

    private boolean isInBound(int[][] board, int x, int y) {
        return x >= 0 && x < board.length && y >= 0 && y < board[0].length;
    }
}