[LeetCode]289. Game of Life
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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):
- 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) of the board given its current state.
Follow up:
- 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?
To solve it in place, we use 2 bits to store 2 states:
[2nd bit, 1st bit] = [next state, current state]- 00 dead (next) <- dead (current)- 01 dead (next) <- live (current) - 10 live (next) <- dead (current) - 11 live (next) <- live (current)
- In the beginning, every cell is either
00
or01
. - Notice that
1st
state is independent of2nd
state. - Imagine all cells are instantly changing from the
1st
to the2nd
state, at the same time. - Let's count # of neighbors from
1st
state and set2nd
state bit. - Since every
2nd
state is by default dead, no need to consider transition01 -> 00
. - In the end, delete every cell's
1st
state by doing>> 1
.
For each cell's 1st
bit, check the 8 pixels around itself, and set the cell's 2nd
bit.
- Transition
01 -> 11
: whenboard == 1
andlives >= 2 && lives <= 3
. - Transition
00 -> 10
: whenboard == 0
andlives == 3
.
To get the current state, simply do
board[i][j] & 1
To get the next state, simply do
board[i][j] >> 1
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 the 2nd bit will 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;}
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