[leetcode] 289. Game of Life 解题报告

题目链接: https://leetcode.com/problems/game-of-life/

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?

思路: 如果可以开额外的空间将会很简单, 就是计算一下周围活的有几个, 然后设置一下次状态. 如果就地的话也可以设置一个另外的标记, 就是如果原来是死的, 下一次活了, 就将其设置为2; 如果当前是活的下次死了, 那就设置为3. 最后在将2, 3的标记分别设为1, 0就可以了.

代码如下:

class Solution {public:    void gameOfLife(vector<vector<int>>& board) {        if(board.size()==0 || board[0].size()==0) return;        int m = board.size(), n = board[0].size();        for(int i = 0; i< m ; i++)        {            for(int j =0; j< n; j++)            {                int s1 = 0;                if(i-1>=0&&(board[i-1][j]==1||board[i-1][j]==3)) s1++;                if(i-1>=0&&j-1>= 0&&(board[i-1][j-1]==1||board[i-1][j-1]==3)) s1++;                if(i-1>=0&&j+1<n&&(board[i-1][j+1]==1||board[i-1][j+1]==3)) s1++;                if(i+1<m && (board[i+1][j]==1 || board[i+1][j]==3)) s1++;                if(i+1<m&&j-1>=0&&(board[i+1][j-1]==1||board[i+1][j-1]==3)) s1++;                if(i+1<m&&j+1<n&&(board[i+1][j+1]==1||board[i+1][j+1]==2)) s1++;                if(j-1>=0&&(board[i][j-1]==1 || board[i][j-1]==3)) s1++;                if(j+1 < n && (board[i][j+1] == 1 || board[i][j+1]==3)) s1++;                    if(board[i][j]==1 && (s1 < 2||s1>3)) board[i][j] = 3;                else if(board[i][j]==0 && s1 == 3) board[i][j] = 2;            }        }        for(int i = 0; i< m; i++)            for(int j =0; j< n; j++)                if(board[i][j] == 2) board[i][j] = 1;                else if(board[i][j] == 3) board[i][j] = 0;    }};