289. Game of Life

Given aboardwithmbyncells, each cell has an initial statelive(1) ordead(0). Each cell interacts with itseight 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?

follow  up 的问题关键在于怎么区别是原来就是活着或者死的， 还是被周围影响的变成活着或者死的。
解决的办法是借用两个值分别表示 0-->1   还是 1-->0 , 1-->1 没有任何变化，可以忽略这个条件。