LeetCode No37. 解數獨
題目
編寫一個程式，通過填充空格來解決數獨問題。
數獨的解法需 遵循如下規則：
數字 1-9 在每一行只能出現一次。
數字 1-9 在每一列只能出現一次。
數字 1-9 在每一個以粗實線分隔的 3x3 宮內只能出現一次。（請參考示例圖）
數獨部分空格內已填入了數字，空白格用 '.' 表示。
示例 1：
輸入：board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]]
輸出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解釋：輸入的數獨如上圖所示，唯一有效的解決方案如下所示：
提示：
board.length == 9
board[i].length == 9
board[i][j] 是一位數字或者 '.'
題目資料 保證 輸入數獨僅有一個解
思路
其實，如果知道上一題的雜湊解法，這道題目就很簡單了，回溯直接解決。
AC程式碼
點選檢視程式碼
class Solution { boolean[][] rowOk = new boolean[10][10]; boolean[][] colOk = new boolean[10][10]; boolean[][][] littleSudo = new boolean[3][3][10]; boolean end = false; List<Integer> blankPos = new ArrayList<Integer>(); public void DFS(char[][] board, int cur) { if( cur==blankPos.size() ) { end = true; return ; } int pos = blankPos.get(cur); int i = pos/10; int j = pos%10; for(int num=1; num<10 && !end; num++) { if( !rowOk[i][num] && !colOk[j][num] && !littleSudo[i/3][j/3][num] ) { rowOk[i][num] = true; colOk[j][num] = true; littleSudo[i/3][j/3][num] = true; board[i][j] = (char) (num+'0'); DFS(board, cur+1); rowOk[i][num] = false; colOk[j][num] = false; littleSudo[i/3][j/3][num] = false; } } } public void solveSudoku(char[][] board) { for(int i=0; i<9; i++) { for(int j=0; j<9; j++) { if( board[i][j] == '.' ) { blankPos.add(i*10+j); } else { int num = (int) (board[i][j]-'0'); rowOk[i][num] = true; colOk[j][num] = true; littleSudo[i/3][j/3][num] = true; } } } DFS(board, 0); } }