【数组】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?

思路：

这里我们假设对于某个点，值的含义为

0 : 上一轮是0，这一轮过后还是0
1 : 上一轮是1，这一轮过后还是1
2 : 上一轮是1，这一轮过后变为0
3 : 上一轮是0，这一轮过后变为1

这样，对于一个节点来说，如果它周边的点是1或者2，就说明那个点上一轮是活的。最后，在遍历一遍数组，把我们编码再解回去，即0和2都变回0，1和3都变回1，就行了。

/**
 * @param {number[][]} board
 * @return {void} Do not return anything, modify board in-place instead.
 */
var gameOfLife = function(board) {
    var m=board.length,n=board[0].length;
    
    for(var i=0;i<m;i++){
        for(var j=0;j<n;j++){
            var lives=0;
            if(i>0){
                lives+=board[i-1][j]==1||board[i-1][j]==2?1:0;
            }
            if(i<m-1){
                lives+=board[i+1][j]==1||board[i+1][j]==2?1:0;
            }
            if(j>0){
                lives+=board[i][j-1]==1||board[i][j-1]==2?1:0;
            }
            if(j<n-1){
                lives+=board[i][j+1]==1||board[i][j+1]==2?1:0;
            }
            if(i>0&&j>0){
                lives+=board[i-1][j-1]==1||board[i-1][j-1]==2?1:0;
            }
            if(i>0&&j<n-1){
                lives+=board[i-1][j+1]==1||board[i-1][j+1]==2?1:0;
            }
            if(i<m-1&&j>0){
                lives+=board[i+1][j-1]==1||board[i+1][j-1]==2?1:0;
            }
            if(i<m-1&&j<n-1){
                lives+=board[i+1][j+1]==1||board[i+1][j+1]==2?1:0; 
            }
            if(board[i][j] == 0 && lives == 3){
                board[i][j] = 3;
            } else if(board[i][j] == 1){
                if(lives < 2 || lives > 3) 
                    board[i][j] = 2;
            }
        }
    }
    
     for(var i = 0; i < m; i++){
        for(var j = 0; j < n; j++){
            board[i][j] = board[i][j] % 2;
        }
    }
};