LeetCode

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?

在原空间上改变是这个题的困难之处，因为改变后会对下一个数的状态产生影响。而原数组每个数大小只占1位，所以可以利用原数组的其他位来存数。

- 00  dead (next) <- dead (current)- 01  dead (next) <- live (current)  - 10  live (next) <- dead (current)  - 11  live (next) <- live (current) 

这样做出改变后，只要在最后将数组所有数都右移一位就好了。时间复杂度O(nm)

class Solution {public:    void gameOfLife(vector<vector<int>>& board) {        int n = board.size();        int m = board[0].size();        for (int i = 0; i < n; ++i) {            for (int j = 0; j < m; ++j) {                int cnt = 0;                for (int k1 = max(0, i-1); k1 <= min(n-1, i+1); ++k1) {                    for (int k2 = max(0, j-1); k2 <= min(m-1, j+1); ++k2) {                        cnt += board[k1][k2] & 1;                    }                }                cnt -= board[i][j] & 1;                if (board[i][j] && (cnt == 2 || cnt == 3))                    board[i][j] = (1 << 1) + 1;                else if (!board[i][j] && cnt == 3)                     board[i][j] = (1 << 1) + 0;            }        }        for (int i = 0; i < n; ++i) {            for (int j = 0; j < m; ++j) {                board[i][j] >>= 1;            }        }        return;    }};