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:
References:点这里
这道题思路很明确(遍历+更新),但是题目明确告诉我们,需要in-place,那么我们只能在原数组里操作,而一些中间状态的处理是难点。下面的解法还是很好的。
To solve it in place, we use 2 bits to store 2 states:
[2nd bit, 1st bit] = [next state, current state]
- 00 dead (current) -> dead (next)
- 01 live (current) -> dead (next)
- 10 dead (current) -> live (next)
- 11 live (current) -> live (next)
In the beginning, every 2nd state is 0
; when next becomes alive change 2nd bit to 1
:
nbs < 2 || nbs > 3
(we don't need to care!)nbs >= 2 && nbs <= 3
nbs == 3
To get this state, we simple do:
board[i][j] & 1
To get next state, we simple do:
board[i][j] >> 1
Hope this helps!
在下面的核心for循环里,只处理了11和01两种情况,因为01和00会在board[i][j] >> 1的时候直接处理掉。
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 2nd bit will become 1. if((board[i][j] & 1) == 1 && (lives >= 2 && lives <= 3)) { board[i][j] = 3; // Make the 2nd bit 1: 01 ---> 11 } if((board[i][j] & 1) == 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 p = Math.max(i - 1, 0); p <= Math.min(i + 1, m - 1); p++) { for(int q = Math.max(j - 1, 0); q <= Math.min(j + 1, n - 1); q++) { lives += board[p][q] & 1; } } lives -= board[i][j] & 1; return lives; }