LeetCode: 289. Game of Life

友情支持

如果您觉得这个笔记对您有所帮助，看在D瓜哥码这么多字的辛苦上，请友情支持一下，D瓜哥感激不尽，😜

支付宝

有些打赏的朋友希望可以加个好友，欢迎关注D 瓜哥的微信公众号，这样就可以通过公众号的回复直接给我发信息。

wx jikerizhi

公众号的微信号是: jikerizhi。因为众所周知的原因，有时图片加载不出来。如果图片加载不出来可以直接通过搜索微信号来查找我的公众号。

289. 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):

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. 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:

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.
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?

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
/**
 * Runtime: 0 ms, faster than 100.00% of Java online submissions for Game of Life.
 *
 * Memory Usage: 35.1 MB, less than 100.00% of Java online submissions for Game of Life.
 */
public void gameOfLife(int[][] board) {
    if (Objects.isNull(board) || board.length == 0 || board[0].length == 0) {
        return;
    }
    for (int i = 0; i < board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            board[i][j] = countOfLive(board, i, j);
        }
    }
    for (int i = 0; i < board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            int value = 0;
            if (isLive(board, i, j)) {
                value = 1;
            }
            board[i][j] = value;
        }
    }
}

private boolean isLive(int[][] matrix, int i, int j) {
    int value = matrix[i][j];
    if (value == 12 || value == 13 || value == -3) {
        return true;
    } else {
        return false;
    }
}

private int countOfLive(int[][] matrix, int i, int j) {
    int count = 0;
    int value = matrix[i][j];
    int iLength = matrix.length;
    int jLength = matrix[0].length;


    if (i - 1 >= 0) {
        if (j - 1 >= 0) {
            count += matrix[i - 1][j - 1] > 0 ? 1 : 0;
        }

        count += matrix[i - 1][j] > 0 ? 1 : 0;

        if (j + 1 < jLength) {
            count += matrix[i - 1][j + 1] > 0 ? 1 : 0;
        }
    }

    if (j - 1 >= 0) {
        count += matrix[i][j - 1] > 0 ? 1 : 0;
    }
    if (j + 1 < jLength) {
        count += matrix[i][j + 1] > 0 ? 1 : 0;
    }

    if (i + 1 < iLength) {
        if (j - 1 >= 0) {
            count += matrix[i + 1][j - 1] > 0 ? 1 : 0;
        }

        count += matrix[i + 1][j] > 0 ? 1 : 0;

        if (j + 1 < jLength) {
            count += matrix[i + 1][j + 1] > 0 ? 1 : 0;
        }
    }

    if (value > 0) {
        return count + 10;
    } else {
        return -count;
    }
}