LeetCode 289. Game of Life (C++)

题目：

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. The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously.

Example:

Input: 
[
  [0,1,0],
  [0,0,1],
  [1,1,1],
  [0,0,0]
]
Output: 
[
  [0,0,0],
  [1,0,1],
  [0,1,1],
  [0,1,0]
]

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：

矩阵中每个元素为1（live）或0（dead），题目给出了细胞更新的规则，求出更新后的矩阵。规则如下：

1. 如果活细胞周围八个位置的活细胞数少于两个，则该位置活细胞死亡；
2. 如果活细胞周围八个位置有两个或三个活细胞，则该位置活细胞仍然存活；
3. 如果活细胞周围八个位置有超过三个活细胞，则该位置活细胞死亡；
4. 如果死细胞周围正好有三个活细胞，则该位置死细胞复活；

首先先判断元素是0还是1，再计算周围八个元素的值的和，根据规则更新。创建一个新的二维数组，最后将计算的值赋给原数组。

程序：

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        vector<vector<int>> res;
        vector<int> temp;
        for (int i = 0; i < board.size(); i++){
            for (int j = 0; j < board[0].size(); j++){
                //dead cell
                if (board[i][j] == 0){
                    int alive = 0;
                    for (int m = i-1; m <= i+1; m++){
                        for (int n = j-1; n <= j+1; n++){
                            if (m < 0 || m >= board.size() || n < 0 || n >= board[0].size())
                                continue;
                            else{
                                if (board[m][n] == 1)
                                    alive++;
                            }
                        }
                    }
                    if (alive == 3)
                        temp.push_back(1);
                    else
                        temp.push_back(0);
                }
                //live cell
                else{
                    int al = 0;
                    for (int m = i-1; m <= i+1; m++){
                        for (int n = j-1; n <= j+1; n++){
                            if (m < 0 || m >= board.size() || n < 0 || n >= board[0].size())
                                continue;
                            else{
                                if (board[m][n] == 1)
                                    al++;
                            }
                        }
                    }
                    //计算了board[m][n]的值，所以判断条件+1
                    if (al < 3)
                        temp.push_back(0);
                    else if (al > 4)
                        temp.push_back(0);
                    else
                        temp.push_back(1);
                }
            }
            res.push_back(temp);
            temp.clear();
        }
        for (int i = 0; i < board.size(); i++){
            for (int j = 0; j < board[0].size(); j++){
                board[i][j] = res[i][j];
            }
        }
    }
};

备注：

做法是新开辟了一个数组，以后再更新下用原数组的程序。