文章目錄

• • 1. 題目
  • 2. 解題
  • • 2.1 復制數組解法
    • 2.2 原地解法
    • 2.3 位運算

1. 題目

鏈接：289. 生命游戲

2. 解題

2.1 復制數組解法

很簡單，按照題意模擬即可

class Solution { public:void gameOfLife(vector<vector<int>>& board) {if(board.empty())return;vector<vector<int>> cp(board);//復制數組法int nearby[8][2] = {{-1, 0}, {-1, -1}, {0, -1}, {1, -1},{1, 0}, {1, 1}, {0, 1}, {-1, 1}};int m = board.size(), n = board[0].size(), i, j, k, x, y, live;for(i = 0; i < m; ++i){for(j = 0; j < n; ++j){live = 0;for(k = 0; k < 8; ++k){x = i+nearby[k][0];y = j+nearby[k][1];if(x>=0 && x<m && y>=0 && y<n && cp[x][y])live++;}if(cp[i][j]){if(live < 2 || live > 3)board[i][j] = 0;}else{if(live == 3)board[i][j] = 1;}}}} };

2.2 原地解法

• 原來是1，標記成0，用2表示（等于1或者2，原狀態都是1）
• 原來是0，標記成1，用3表示（等于0或者3，原狀態都是0）
• 處理完后，改寫數組
• 等于2的，賦值為0
• 等于3的，賦值為1

class Solution { public:void gameOfLife(vector<vector<int>>& board) {if(board.empty())return;int nearby[8][2] = {{-1, 0}, {-1, -1}, {0, -1}, {1, -1},{1, 0}, {1, 1}, {0, 1}, {-1, 1}};int m = board.size(), n = board[0].size(), i, j, k, x, y, live;for(i = 0; i < m; ++i){for(j = 0; j < n; ++j){live = 0;for(k = 0; k < 8; ++k){x = i+nearby[k][0];y = j+nearby[k][1];if(x>=0 && x<m && y>=0 && y<n && (board[x][y]==1 || board[x][y]==2))live++;}if(board[i][j]){if(live < 2 || live > 3)board[i][j] = 2;}else{if(live == 3)board[i][j] = 3;}}}for(i = 0; i < m; ++i){for(j = 0; j < n; ++j){if(board[i][j] == 2)board[i][j] = 0;else if(board[i][j] == 3)board[i][j] = 1;}}} };

2.3 位運算

• 利用數值的低位作為原數據
• 高位保存周圍存活的數量

class Solution { public:void gameOfLife(vector<vector<int>>& board) {if (board.empty())return;int m = board.size(), n = board[0].size(), i, j, k, x, y, live;int nearby[8][2] = {{-1, 0}, {-1, -1}, {0, -1}, {1, -1},{1, 0}, {1, 1}, {0, 1}, {-1, 1}}; for (i = 0; i < m; ++i) {for (j = 0; j < n; ++j) {for (k = 0; k < 8; ++k) {x = i + nearby[k][0];y = j + nearby[k][1];if (x >= 0 && x < m && y >= 0 && y < n)board[i][j] += (board[x][y] & 1) << 1;}}}for (i = 0; i < m; ++i) {for (j = 0; j < n; ++j) {live = board[i][j] >> 1;if (live < 2 || live > 3) board[i][j] = 0;else if (live == 3)board[i][j] = 1;elseboard[i][j] &= 1;}}} };

總結

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