[leetcode] 289. Game of Life

According to theWikipedia's article: "TheGame of Life, also known simply asLife, is a cellular automaton devised by the British mathematician John Horton Conway in 1970."

Given aboard withm 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?

这道题是小游戏“Game of Life”，给出当前状态让求出下一状态，题目难度为Medium。

根据题目的描述这道题不难，关键在于“in-place”。这里board中记录状态用的是int，但其实只用了1bit，其他31位还是可以利用的，想到这里问题也就没什么难度了，不过这能算得上真正的in-place吗？看了看别人的代码，似乎也没有其他的方法，如果这里board中数据从int变为bool该怎么处理呢？有更好方法的朋友可以给我留言。

这里用第二位来存储下一状态，这样最后只需要遍历board，把其中状态数据右移一位就完成了状态变更。具体代码：

class Solution {    int getLiveNbCnt(vector<vector<int>>& board, int m, int n, int row, int col) {        int cnt = 0;        for(int i=max(row-1, 0); i<=min(row+1, m-1); i++) {            for(int j=max(col-1, 0); j<=min(col+1, n-1); j++) {                cnt += (board[i][j]&0x1);            }        }        return cnt-board[row][col];    }public:    void gameOfLife(vector<vector<int>>& board) {        if(board.empty()) return;        int m = board.size();        int n = board[0].size();        int nbCnt = 0;                for(int i=0; i<m; i++) {            for(int j=0; j<n; j++) {                nbCnt = getLiveNbCnt(board, m, n, i, j);                if(board[i][j]&0x1 && nbCnt>=2 && nbCnt<=3) {                    board[i][j] |= 2;                }                else if((board[i][j]&0x1)==0 && nbCnt==3) {                    board[i][j] |= 2;                }            }        }                for(int i=0; i<m; i++) {            for(int j=0; j<n; j++) {                board[i][j] >>= 1;            }        }    }};