289. Game of Life

Description

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Solution

注意在边界上的cell，neighbor数量会不足8个，不用跨到另一边去寻找。

Iterative, time O(m * n), space O(m * n)

class Solution {
    public void gameOfLife(int[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) {
            return;
        }
        
        int m = board.length;
        int n = board[0].length;
        int[] next = new int[m * n];
        
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int liveCount 
                    = getNeighbor(board, i - 1, j - 1, m, n)
                    + getNeighbor(board, i - 1, j, m, n)
                    + getNeighbor(board, i - 1, j + 1, m, n)
                    + getNeighbor(board, i, j - 1, m, n)
                    + getNeighbor(board, i, j + 1, m, n)
                    + getNeighbor(board, i + 1, j - 1, m, n)
                    + getNeighbor(board, i + 1, j, m, n)
                    + getNeighbor(board, i + 1, j + 1, m, n);
                
                if ((board[i][j] == 1 && liveCount >= 2 && liveCount <= 3)
                   || (board[i][j] == 0 && liveCount == 3)) {
                    next[i * n + j] = 1;
                } else {
                    next[i * n + j] = 0;
                }
            }
        }
        
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                board[i][j] = next[i * n + j];
            }
        }
    }
    
    public int getNeighbor(int[][] board, int i, int j, int m, int n) {
        if (i < 0 || i >= m || j < 0 || j >= n) {
            return 0;
        }
        return board[i][j];
    }
}

Bit manipulation, time O(m * n), space O(1)

由于board里面的值非0即1，可以用额外的二进制位去存储下一个state。将下一个state存储在倒数第二位就可以了呀。

注意这里getNeighbors的方式也很值得借鉴，比上面的方式简洁很多。

class Solution {
    public void gameOfLife(int[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) {
            return;
        }
        
        int m = board.length;
        int n = board[0].length;
        
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int liveCount = getLiveNeighbors(board, i, j, m, n);
                
                if ((board[i][j] == 1 && liveCount >= 2 && liveCount <= 3)
                   || (board[i][j] == 0 && liveCount == 3)) {
                    board[i][j] |= 2;   // save next state in the 31st bit
                }
            }
        }
        
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                board[i][j] >>= 1;  // get next state from the 31st bit
            }
        }
    }
    
    public int getLiveNeighbors(int[][] board, int i, int j, int m, int n) {
        int lives = 0;
        
        for (int x = Math.max(i - 1, 0); x <= Math.min(i + 1, m - 1); ++x) {
            for (int y = Math.max(j - 1, 0); y <= Math.min(j + 1, n - 1); ++y) {
                if (x == i && y == j) {
                    continue;
                }
                lives += board[x][y] & 1;   // get current state from the 32nd bit
            }
        }
        
        return lives;
    }
}