# LeetCode – Game of Life (Java)

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:

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.

**Java Solution 1**

Because we need to solve the problem in place, we can use the higher bit to record the next state. And at the end, shift right a bit to get the next state for each cell.

public void gameOfLife(int[][] board) { if(board==null || board.length==0||board[0].length==0) return; int m=board.length; int n=board[0].length; int[] x = {-1, -1, 0, 1, 1, 1, 0, -1}; int[] y = {0, 1, 1, 1, 0, -1, -1, -1}; for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ int count=0; for(int k=0; k<8; k++){ int nx=i+x[k]; int ny=j+y[k]; if(nx>=0&&nx<m&&ny>=0&&ny<n&&(board[nx][ny]&1)==1){ count++; } } //<2 die if(count<2){ board[i][j] &= 1; } //same state if(count==2||count==3){ board[i][j] |= board[i][j]<<1; } //go live if(count==3){ board[i][j] |=2; } //>3 die if(count>3){ board[i][j] &=1; } } } for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ board[i][j] = board[i][j]>>1; } } } |

<pre><code> String foo = "bar"; </code></pre>