标签：ems initial 方向 art length repr and after nbsp
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):
Write a function to compute the next state (after one update) of the board given its current state.
Follow up:
解题思路：因为要求in-place,而状态转换最多也就四种，且只要记录两种即0->1,1->0这两种即可。于是可以现将board的值置为-1，-2，分别赋予两种状态转换的含义，最后再把负数换成本要转换的0，1状态即可。
class Solution { public: //-1: 1->0 -2:0->1 void countNeighbors(vector<vector<int>>& board, int x, int y, int& nLive) { int n = board.size(), m = board[0].size(); int dir[8][2] = { {-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1,1} }; for(int i = 0; i < 8; ++i) { int xx = x + dir[i][0]; int yy = y + dir[i][1]; if(xx >= 0 && xx < n && yy >= 0 && yy < m) { if(board[xx][yy] == 1 || board[xx][yy] == -1) nLive++; } } } void gameOfLife(vector<vector<int>>& board) { if(board.empty()) return; int n = board.size(), m = board[0].size(), nLive; for(int i = 0; i < n; ++i) { for(int j = 0; j < m; ++j) { nLive=0; countNeighbors(board, i, j, nLive); if(board[i][j] == 1) { if(nLive < 2 || nLive > 3) board[i][j] = -1; } else { if(nLive == 3) board[i][j] = -2; } } } for(int i = 0; i < n; ++i) for(int j = 0; j < m; ++j) { if(board[i][j] == -1) board[i][j] = 0; else if(board[i][j] == -2) board[i][j] = 1; } } };
看了discuss里的，用低一位表示初始状态，高一位表示改变之后的状态，这样最后只要去掉初始状态就能得到最终状态。ps：八个方向搜索可以借鉴下：
public void gameOfLife(int[][] board) { if (board == null || board.length == 0) return; int m = board.length, n = board[0].length; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { int lives = liveNeighbors(board, m, n, i, j); // In the beginning, every 2nd bit is 0; // So we only need to care about when will the 2nd bit become 1. if (board[i][j] == 1 && lives >= 2 && lives <= 3) { board[i][j] = 3; // Make the 2nd bit 1: 01 ---> 11 } if (board[i][j] == 0 && lives == 3) { board[i][j] = 2; // Make the 2nd bit 1: 00 ---> 10 } } } for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { board[i][j] >>= 1; // Get the 2nd state. } } } public int liveNeighbors(int[][] board, int m, int n, int i, int j) { int lives = 0; for (int x = Math.max(i - 1, 0); x <= Math.min(i + 1, m - 1); x++) { for (int y = Math.max(j - 1, 0); y <= Math.min(j + 1, n - 1); y++) { lives += board[x][y] & 1; } } lives -= board[i][j] & 1; return lives; }
标签：ems initial 方向 art length repr and after nbsp
原文地址：http://www.cnblogs.com/tsunami-lj/p/7634588.html
