LeetCode 289 Game of Life
2015-12-14 10:15
363 查看
题目描述
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.
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?
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
分析
参考:leetcode 289: Game of Life(西施豆腐渣博客)题目中说最好是in-place,因为board是int型的,所以要区分四种状态(dead是0,live是1):
[code]dead->live 01 dead->dead 00 live->dead 10 live->live 11
或者直接用0,1,2,3四个数字表示四种状态都可以,然后通过移位进行结果的更新。
下面的图片,是一个Game of Life变化图。
代码
[code] public void gameOfLife(int[][] board) { if (board == null || board.length == 0 || board[0] == null || board[0].length == 0) return; int m = board.length; int n = board[0].length; // 判断每个点,下一时刻的状态 for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { int x = getLiveNum(board, i, j); if (board[i][j] == 0) { if (x == 3) board[i][j] += 10; } else { if (x == 2 || x == 3) board[i][j] += 10; } } } // 更新每个点的状态 for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { board[i][j] /= 10; } } } // 获取board[x][y]邻居live的数量 private int getLiveNum(int[][] board, int x, int y) { int c = 0; for (int i = x - 1; i <= x + 1; i++) { for (int j = y - 1; j <= y + 1; j++) { if (i < 0 || j < 0 || i > board.length - 1 || j > board[0].length - 1 || (i == x && j == y)) continue; if (board[i][j] % 10 == 1) c++; } } return c; }
相关文章推荐
- JS中Json对象和字符串之间的转换
- 前端开发工程化探讨--基础篇(长文)
- JSON使用DataContract和DataMember
- 前端中实现页内跳转和导航栏点击实现样式,点击其他消除样式并为这一项添加样式
- 浅谈Ajax
- jquery easyui结合mysql数据库实现后台分页
- React-Native环境配置
- javascript script load related【转】
- css命名书写规范小结。
- HTML标签与表格
- JSON 的标准:双引号而非单引号!
- 【caffe跑试验遇到错误:Check failed: error == cudaSuccess (2 vs. 0) out of memory】
- 用CNZZ实现Html5中的事件统计
- jquery获取所有选中的checkbox
- React-Native植入原声应用 - Android
- 一天JavaScript示例-在功能上的标量参数和数组参数的差异
- jquery结合Spring MVC实现从后台读取数据的输入框提示
- JSP自定义标签开发入门
- 给产品经理讲技术丨前端和后台的数据交互与协议
- 日常笔记之Buffer的拼接