# 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):

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]
]
``````

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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

``````class Solution {
public:
void gameOfLife(vector<vector<int> >& board) {
int m = board.size(), n = m ? board[0].size() : 0;
vector<int> dx{-1, -1, -1, 0, 1, 1, 1, 0};
vector<int> dy{-1, 0, 1, 1, 1, 0, -1, -1};
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int cnt = 0;
for (int k = 0; k < 8; ++k) {
int x = i + dx[k], y = j + dy[k];
if (x >= 0 && x < m && y >= 0 && y < n && (board[x][y] == 1 || board[x][y] == 2)) {
++cnt;
}
}
if (board[i][j] && (cnt < 2 || cnt > 3)) board[i][j] = 2;
else if (!board[i][j] && cnt == 3) board[i][j] = 3;
}
}
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
board[i][j] %= 2;
}
}
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/289

Set Matrix Zeroes

https://leetcode.com/problems/game-of-life/

https://leetcode.com/problems/game-of-life/discuss/73217/Infinite-board-solution

https://leetcode.com/problems/game-of-life/discuss/73230/C%2B%2B-O(1)-space-O(mn)-time

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

LeetCode All in One 题目讲解汇总(持续更新中…)

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