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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?
题意：生命游戏。。。看到follow up本以为是提高篇，结果不是题目要求。可是第一个又说 board要同时更新，这是要求吧。于是我就按照棋盘是二维环来写的，最后发现棋盘是有限的。
也就是两个follow up，要看1（同时变换），不要看2（实际上棋盘是有限的）。这不是有毒吗？
所以做leetcode最大的挑战是如何保持心平气和，不要被傻逼题目以及傻逼题目描述影响心情（
我直接写了棋盘是二维环的版本。注意的就是，八个方向遍历后记得去重，以及不能和起点相同。
同时更新再加两个标记就好了。
1代表活的，0代表死了
-1代表刚才活的，现在死了
-2代表刚才死的，现在活了。
最后记得再变回来。。。
代码实现
1/* ***********************************************
2Author :111qqz
3Created Time :2017年04月05日 星期三 18时52分29秒
4File Name :289.cpp
5************************************************ */
6class Solution {
7
8public:
9 const int dx[8]={1,1,1,0,0,-1,-1,-1};
10 const int dy[8]={-1,0,1,-1,1,-1,0,1};
11 const int N=1E3+3;
12 int n,m;
13 int checkX( int x)
14 {
15 if (x>=0&&x<=n-1) return x;
16 if (x==n) return 0;
17 if (x==-1) return n-1;
18 }
19 int checkY( int y)
20 {
21 if (y>=0&&y<=m-1) return y;
22 if (y==m) return 0;
23 if (y==-1) return m-1;
24 }
25 bool check ( int x,int y) //follow up 里的不是题意啊。。。有毒。。。
26 {
27 if (x>=0&&x<=n-1&&y>=0&&y<=m-1) return true;
28 return false;
29 }
30
31 int live( int x,int y,vector<vector<int> >& maze)
32 {
33 int cnt = 0 ;
34 set < pair < int ,int > > se;
35
36 for ( int i = 0 ; i < 8 ; i++) //八个方向经过check以后可能有相同的orz...
37 { //还可能遇到自己orz...
38 int nx = x + dx[i];
39 int ny = y + dy[i];
40 // nx = checkX(nx);
41 // ny = checkY(ny);
42 if (!check(nx,ny)) continue;
43 if (nx==x&&ny==y) continue;
44// cout<<"nx:"<<nx<<" ny:"<<ny<<" maze:"<<maze[nx][ny]<<endl;
45 if (maze[nx][ny]==1||maze[nx][ny]==-1) se.insert(make_pair(nx,ny));
46 }
47 cnt = se.size();
48 //cout<<"x:"<<x<<" y:"<<y<<" cnt:"<<cnt<<endl;
49 if (cnt<2||cnt>3) return -1;
50 else
51 return 1;
52
53 } //1 -> live ,0->dead
54 int dead( int x,int y,vector<vector<int> >& maze) //-1>刚才活的，现在死了， -2->刚才死的，现在活了
55 {
56 int cnt = 0 ;
57 set< pair <int,int > >se;
58 for ( int i = 0 ; i < 8 ; i++)
59 {
60 int nx = x + dx[i];
61 int ny = y + dy[i];
62 //nx = checkX(nx);
63 //ny = checkY(ny);
64 if (!check(nx,ny)) continue;
65 if (nx==x&&ny==y) continue;
66// printf("nx:%d ny:%d\n",nx,ny);
67 if (maze[nx][ny]==1||maze[nx][ny]==-1) se.insert(make_pair(nx,ny));
68 }
69 cnt = se.size();
70 if (cnt==3) return -2; //死的变活了
71 return 0;
72
73 }
74 void gameOfLife(vector<vector<int>>& board) {
75 n = board.size();
76 if (n==0) return;
77 m = board[0].size();
78 if (m==0) return;
79 for ( int i = 0 ; i < n ; i++)
80 {
81 for ( int j = 0 ;j < m ; j++)
82 {
83 int tmp = board[i][j];
84// cout<<"val:"<<tmp<<endl;
85 if (tmp==1||tmp==-1)
86 {
87 board[i][j] = live(i,j,board);
88 }
89 else if (tmp==0||tmp==-2)
90 {
91 board[i][j] = dead(i,j,board);
92 }
93 }
94 }
95 for ( int i = 0 ;i < n ; i++)
96 for ( int j = 0 ; j < m ; j++)
97 if (board[i][j]==-2) board[i][j] = 1;
98 else if (board[i][j]==-1) board[i][j] = 0;
99
100 }
101
102};