[Java知识库]Leetcode--Java--37. 解数独

题目描述

编写一个程序，通过填充空格来解决数独问题。

数独的解法需 遵循如下规则：

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）
数独部分空格内已填入了数字，空白格用 ‘.’ 表示。

样例描述

输入：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"]]
输出：[["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"]]

思路

1. 递归回溯法。
   
   对于每一个空位，根据布尔数组看每一行、列，九宫格哪些数被填过了，枚举所有可以填的数字。然后继续，如果都不能填，就回溯到上一个状态。

代码

class Solution {

         //标记row[i][t]第i行 的是否存在t
      boolean  row[][] = new boolean[9][9];
      boolean  col[][] = new boolean[9][9];
      //cell通过i/3,j/3坐标来确定存在数字t
      boolean cell[][][] = new boolean[3][3][9];

    public void solveSudoku(char[][] board) {

      //初始化board已经有的数
      for (int i = 0; i < 9; i ++ ) {
          for (int j = 0; j < 9; j ++ ) {
              if (board[i][j] != '.') {
                 //获取数  减一因为数放在下标0-8
                 int t = board[i][j] - '1';   
                 //将该数标记在对应的行 列以及九宫格
                 row[i][t] = col[j][t] = cell[i / 3][j / 3][t] = true;
              }
          }
      }

      dfs(board, 0, 0);
    }
   //通过返回boolean来判断当前分支是否可以，所以不是void
    public boolean dfs(char[][] board, int x, int y) {
        //对于每一行，从左往右根据列来填
        //这一行所有列填完
        if (y == 9) {
            //换下一行，从第一列开始
            x ++;
            y = 0;
        }
        //所有行填完，x到9 说明数独全部填完
        if (x == 9) return true;
        //如果这个位置已经有数，直接下一个,一定要带return！ 因为要标志分支是否可解
        if (board[x][y] != '.')  return dfs(board, x, y + 1);
        //否则是空位 就开始枚举所有可以填的数
        for (int num = 0; num < 9; num ++ ) {
            //如果这个数不在行、列和九宫格出现过才能填
            if (!row[x][num] && !col[y][num] && !cell[x / 3][y / 3][num]) {
                //填入数   加'1'恢复数
                board[x][y] = (char) (num + '1');
                row[x][num] = col[y][num] = cell[x / 3][y / 3][num] = true;
               //当前填完后，如果下一个也成功，才说明是正确的
               if (dfs(board, x, y + 1)) return true;
               //否则，下一个失败的话。 进行回溯，恢复所有状态
               board[x][y] = '.';
               row[x][num] = col[y][num] = cell[x / 3][y / 3][num] = false;
            }
        }
        //如果上面for循环的都出现在九宫格了，说明当前分支无解，return到上个状态
        return false;
    }
}