【leetcode】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?

翻译：

二维数组的每一个元素代表一个二维空间的一个细胞，0代表死亡，1代表存活，每个细胞与周围8个细胞相邻。每次迭代时，细胞存活与否符合以下四条原则：

1.若一个活细胞，周围的活细胞数小于2个，则细胞转为死亡。（太孤单）

2.若一个活细胞，周围的活细胞数为2或3个，则细胞还是存活。

3.若一个活细胞，周围的活细胞数大于3个，则细胞转为死亡。（太拥挤）

4.若一个死细胞，周围的活细胞数正好为3个，则细胞重生，转为存活。（繁殖）

其他要求：

1.要求直接在原数组上操作，换句话不能额外生成m*n的数组空间，记录中间结果。

2.数组规模无限大。

思路：

一开始我的思路一直在思考是否有合适的更新策略能够逐步将整个数组进行更新，可以做到更新某些细胞时，不影响其他细胞的下一步更新。后来换一种思路，既然只用0和1记录，int的其他位可以用来记录中间结果，只要最后用移位来生成最终的结果就行了

代码：

class Solution {
public:
	void gameOfLife(vector>& board) {
		int m=board.size();
		int n=m?board[0].size():0;
		int count=0;
		for (int i=0;i=0)?i-1:0;I<((i+2<=m)?i+2:i+1);I++)
				{
					for (int J=(j-1>=0)?j-1:0;J<((j+2<=n)?j+2:j+1);J++)
					{
						if((I!=i || J!=j) && board[I][J]&1)
							count++;
					}
				}
				if (count==3 || count+board[i][j]==3)
				{
					board[i][j]|=2;
				}
			}
		}
		for (int i=0;i>1;
		}
	}
};

结果：