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?

解题思路

题目要求就地解决问题,所以不能计算出结果之后马上更新该位置的值。因此我们考虑用十位和个位分别表示下一代的值和当前代的值，计算live成员个数时，我们累加board[i][j] % 10的值，等用十位把所有位置都标记完后统一更新所有位置的值(board[i][j] /= 10)。

实现代码

C++:

// Runtime: 4 msclass Solution {public:    void gameOfLife(vector
     
      
       >& board) {        for (int i = 0; i < board.size(); i++)        {            for (int j = 0; j < board[0].size(); j++)            {                int num = numOfLive(board, i, j);                if (board[i][j] == 1 && (num == 2 || num == 3) ||                    board[i][j] == 0 && num == 3)                {                    board[i][j] += 10;                }            }        }        for (int i = 0; i < board.size(); i++)        {            for (int j = 0; j < board[0].size(); j++)            {                board[i][j] /= 10;            }        }    }private:    int numOfLive(vector
       
        
         > board, int m, int n)    {        int res = 0;        for (int i = m - 1; i <= m + 1; i++)        {            for (int j = n - 1; j <= n + 1; j++)            {                if (i < 0 || j < 0 || i > board.size() - 1 ||                    j > board[0].size() - 1 || (i == m && j == n))                {                    continue;                }                res += board[i][j] % 10;            }        }        return res;    }};
        
       
      
     

Java：

// Runtime: 1 mspublic class Solution {    public void gameOfLife(int[][] board) {        for (int i = 0; i < board.length; i++) {            for (int j = 0; j < board[0].length; j++) {                int num = numOfLive(board, i, j);                if (board[i][j] == 1 && (num == 2 || num == 3) ||                        board[i][j] == 0 && num == 3) {                    board[i][j] += 10;                }            }        }        for (int i = 0; i < board.length; i++) {            for (int j = 0; j < board[0].length; j++) {                board[i][j] /= 10;            }        }    }    private int numOfLive(int[][] board, int m, int n) {        int res = 0;        for (int i = m - 1; i <= m + 1; i++) {            for (int j = n - 1; j <= n + 1; j++) {                if (i < 0 || j < 0 || i > board.length - 1 ||                        j > board[0].length - 1 || (i == m && j == n)) {                    continue;                }                res += board[i][j] % 10;            }        }        return res;    }}