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| 1 | +package com.fghpdf.GameOfLife; |
| 2 | + |
| 3 | +/** |
| 4 | + * @author fghpdf |
| 5 | + * @date 2019/12/25 |
| 6 | + * https://leetcode.com/problems/game-of-life/ |
| 7 | + * https://leetcode.com/problems/game-of-life/discuss/73223/Easiest-JAVA-solution-with-explanation |
| 8 | + * 没懂 |
| 9 | + **/ |
| 10 | +public class Solution { |
| 11 | + public void gameOfLife(int[][] board) { |
| 12 | + if (board == null || board.length == 0) { |
| 13 | + return; |
| 14 | + } |
| 15 | + int m = board.length, n = board[0].length; |
| 16 | + |
| 17 | + for (int i = 0; i < m; i++) { |
| 18 | + for (int j = 0; j < n; j++) { |
| 19 | + int lives = liveNeighbors(board, m, n, i, j); |
| 20 | + |
| 21 | + // In the beginning, every 2nd bit is 0; |
| 22 | + // So we only need to care about when will the 2nd bit become 1. |
| 23 | + if (board[i][j] == 1 && lives >= 2 && lives <= 3) { |
| 24 | + board[i][j] = 3; // Make the 2nd bit 1: 01 ---> 11 |
| 25 | + } |
| 26 | + if (board[i][j] == 0 && lives == 3) { |
| 27 | + board[i][j] = 2; // Make the 2nd bit 1: 00 ---> 10 |
| 28 | + } |
| 29 | + } |
| 30 | + } |
| 31 | + |
| 32 | + for (int i = 0; i < m; i++) { |
| 33 | + for (int j = 0; j < n; j++) { |
| 34 | + board[i][j] >>= 1; // Get the 2nd state. |
| 35 | + } |
| 36 | + } |
| 37 | + } |
| 38 | + |
| 39 | + public int liveNeighbors(int[][] board, int m, int n, int i, int j) { |
| 40 | + int lives = 0; |
| 41 | + for (int x = Math.max(i - 1, 0); x <= Math.min(i + 1, m - 1); x++) { |
| 42 | + for (int y = Math.max(j - 1, 0); y <= Math.min(j + 1, n - 1); y++) { |
| 43 | + lives += board[x][y] & 1; |
| 44 | + } |
| 45 | + } |
| 46 | + lives -= board[i][j] & 1; |
| 47 | + return lives; |
| 48 | + } |
| 49 | +} |
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