【数组】Leetcode编程题解：289. Game of Life Add to List

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题目：

According to the Wikipedia'sarticle: "The Game of Life, alsoknown simply as Life, is a cellularautomaton devised by the British mathematician John Horton Conway in1970."

Given a board with m by ncells, each cell has an initial state live(1) or dead (0). Each cell interactswith itseightneighbors (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) ofthe board given its current state.

要求：

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?

原先的想法是每一次读到一个点，就检查周围的所有点，但是这样不能满足题目的要求，而且时间复杂度较高。于是，参考网上的代码，以下为摘抄：

由于题目中要求所有点的状态需要一次性发生改变，而且不用额外的空间，这是本题的最大难点。

既然需要“就地解决”，我们不妨分析一下borad的特性：board上的元素有两种状态，生（1）和死（0）。这两种状态存在了一个int型里面。所以我们可以有效利用除最低位的其它位，去保存更新后的状态，这样就不需要有额外的空间了。

具体而言，我们可以用最低位表示当前状态，次低位表示更新后状态：

00(0)：表示当前是死，更新后是死；
01(1)：表示当前是生，更新后是死；
10(2)：表示当前是死，更新后是生；
11(3)：表示当前是神，更新后是生。

提交代码：

class Solution {public:    void gameOfLife(vector<vector<int>>& board) {        int row = board.size();int column = board[0].size();        if(row == 0 || column == 0) return;        for(int x = 0; x < row; x++) {        for(int y = 0; y < column; y++) {        int life = getlive(board, row, column, x, y);        if(board[x][y] == 1 && (life == 2 || life == 3))        board[x][y] = 3;        if(board[x][y] == 0 && life == 3)        board[x][y] = 2;}}for(int x = 0; x < row; x++) {        for(int y = 0; y < column; y++) {        board[x][y] >>= 1;        }    }    }private:int getlive(vector<vector<int>>& board, int row, int column, int x, int y) {int num = 0;for(int i = max(x - 1, 0); i < min(x + 2, row); i++) {for(int j = max(y - 1, 0); j < min(y + 2, column); j++) {if(i == x && j == y) continue;else if(board[i][j] % 2 == 1) num++;}}return num;}};

其中min容易出现越界，应该注意。

应用网址：

http://www.cnblogs.com/jdneo/p/5236373.html

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