Game of Life I & II

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?

分析：https://leetcode.com/problems/game-of-life/discuss/73223/Easiest-JAVA-solution-with-explanation

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 or 01.
• Notice that 1st state is independent of 2nd state.
• Imagine all cells are instantly changing from the 1st to the 2nd state, at the same time.
• Let's count # of neighbors from 1st state and set 2nd state bit.
• Since every 2nd state is by default dead, no need to consider transition 01 -> 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: when board == 1 and lives >= 2 && lives <= 3.
• Transition 00 -> 10: when board == 0 and lives == 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 == ) return;
     int m = board.length, n = board[].length;
     for (int i = ; i < m; i++) {
         for (int j = ; 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] ==  && lives >=  && lives <= ) {
                 board[i][j] = ; // Make the 2nd bit 1: 01 ---> 11
             }
             if (board[i][j] ==  && lives == ) {
                 board[i][j] = ; // Make the 2nd bit 1: 00 ---> 10
             }
         }
     }

     for (int i = ; i < m; i++) {
         for (int j = ; j < n; j++) {
             board[i][j] >>= ;  // Get the 2nd state.
         }
     }
 }

 public int liveNeighbors(int[][] board, int m, int n, int i, int j) {
     int lives = ;
     for (int x = Math.max(i - , ); x <= Math.min(i + , m - ); x++) {
         for (int y = Math.max(j - , ); y <= Math.min(j + , n - ); y++) {
             lives += board[x][y] & ;
         }
     }
     lives -= board[i][j] & ;
     return lives;
 }

Game of Life II

In Conway's Game of Life, cells in a grid are used to simulate biological cells. Each cell is considered to be either alive or dead. At each step of the simulation each cell's current status and number of living neighbors is used to determine the status of the cell during the following step of the simulation.

In this one-dimensional version, there are N cells numbered 0 through N-1. The number of cells does not change at any point in the simulation. Each cell i is adjacent to cells i-1 and i+1. Here, the indices are taken modulo N meaning cells 0 and N-1 are also adjacent to eachother. At each step of the simulation, cells with exactly one living neighbor change their status (alive cells become dead, dead cells become alive).

For example, if we represent dead cells with a '0' and living cells with a '1', consider the state with 8 cells: 01100101 Cells 0 and 6 have two living neighbors. Cells 1, 2, 3, and 4 have one living neighbor. Cells 5 and 7 have no living neighbors. Thus, at the next step of the simulation, the state would be: 00011101

 public void solveOneD(int[] board){
     int n = board.length;
     int[] buffer = new int[n];
     // 根据每个点左右邻居更新该节点情况。
     for(int i = ; i < n; i++){
         int lives = board[(i + n + ) % n] + board[(i + n - ) % n];
         if(lives == ){
             buffer[i] = (board[i] + ) % ;
         } else {
             buffer[i] = board[i];
         }
     }
     for(int i = ; i < n; i++){
         board[i] = buffer[i];
     }
 }

 public void solveOneD(int rounds, int[] board){
     int n = board.length;
     for(int i = ; i < n; i++){
         int lives = board[(i + n + ) % n] %  + board[(i + n - ) % n] % ;
         if(lives == ){
             board[i] = board[i] %  + ;
         } else {
             board[i] = board[i];
         }
     }
     for(int i = ; i < n; i++){
         board[i] = board[i] >=  ? (board[i] + ) %  : board[i] % ;
     }
 }