289. Game of Life

题目：

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?

链接： http://leetcode.com/problems/game-of-life/

题解：

生命游戏。题目比较长，用extra array做的话比较简单，但要求in-place的话，我们就要使用一些技巧。这里的方法来自yavinci，他的很多Java解法都既精妙又易读，真的很厉害。 我们用两个bit位来代表当前回合和下回合的board。

00代表当前dead

01代表当前live， next dead

10代表当前dead，next live

11代表当前和next都是live

按照题意对当前cell进行更新，全部更新完毕以后， 需要再遍历一遍整个数组，将board upgrade到下一回合， 就是每个cell >>= 1。

Time Complexity - O(mn)， Space Complexity - O(1)。

public class Solution {
    public void gameOfLife(int[][] board) {           
        if(board == null || board.length == 0) {
            return;
        }
        
        for(int i = 0; i < board.length; i++) {
            for(int j = 0; j < board[0].length; j++) {
                int liveNeighbors = getLiveNeighbors(board, i, j);
                if((board[i][j] & 1) == 1) {
                    if(liveNeighbors >= 2 && liveNeighbors <= 3) {
                        board[i][j] = 3;             // change to "11", still live
                    }                                // else it stays as "01", which will be eleminated next upgrade
                } else {
                    if(liveNeighbors == 3) {
                        board[i][j] = 2;     // change to "10", become live
                    }
                }
            }
        }
        
        for(int i = 0; i < board.length; i++) {
            for(int j = 0; j < board[0].length; j++) {
                board[i][j] >>= 1;
            }
        }
    }
    
    private int getLiveNeighbors(int[][] board, int row, int col) {
        int res = 0;
        for(int i = Math.max(row - 1, 0); i <= Math.min(board.length - 1, row + 1); i++) {
            for(int j = Math.max(col - 1, 0); j <= Math.min(board[0].length - 1, col + 1); j++) {
                res += board[i][j] & 1;
            }
        }
        res -= board[row][col] & 1;
        return res;
    }
}

Reference:

https://leetcode.com/discuss/68352/easiest-java-solution-with-explanation

https://leetcode.com/discuss/61912/c-o-1-space-o-mn-time

https://leetcode.com/discuss/61910/clean-o-1-space-o-mn-time-java-solution

289. Game of Life

原文：http://www.cnblogs.com/yrbbest/p/5040751.html