题目描述

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

• 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。

一个数独。

答案被标成红色。

Note:

• 给定的数独序列只包含数字 1-9 和字符 '.' 。
• 你可以假设给定的数独只有唯一解。
• 给定数独永远是 9x9 形式的。

样例

输入:
[
  ["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"]
]

算法分析

dfs

• 1、预处理出row，col，cell数组，某些元素的位置已经被标识过
• 2、dfs从(0,0)到(8,8)按顺序递归，假设递归到当前位置是(i,j)
  若该位置已经被表示过，则直接跳去下一个位置
  若该位置未被表示过，则往该位置尝试填1 到 9 的数字，并在row，col，cell数组中进行标记，用完记得恢复现场
• 3、当能成功到达最后一个位置，直接返回true，结束递归

时间复杂度

不好分析

Java 代码

class Solution {
    static boolean dfs(char[][] board,boolean[][] row,boolean[][] col,boolean[][][] cell,int x,int y)
    {
        //判断边界
        if(y == 9) 
        {
            x ++;
            y = 0;
            if(x == 9) return true;

        }
        if(board[x][y] != '.') return dfs(board,row,col,cell,x,y + 1);

        //尝试填数
        for(int i = 1;i <= 9;i ++)
        {
            if(row[x][i] || col[y][i] || cell[x / 3][y / 3][i]) continue;
            board[x][y] = (char)('0' + i);
            row[x][i] = col[y][i] = cell[x / 3][y / 3][i] = true;
            if(dfs(board,row,col,cell,x,y + 1)) return true;
            board[x][y] = '.';
            row[x][i] = col[y][i] = cell[x / 3][y / 3][i] = false;
        }
        return false;
    }
    public void solveSudoku(char[][] board) {
        boolean[][] row = new boolean[10][10];
        boolean[][] col = new boolean[10][10];
        boolean[][][] cell = new boolean[3][3][10];

        //初始化
        for(int i = 0;i < 9;i ++)
            for(int j = 0;j < 9;j ++)
            {
                if(board[i][j] != '.')
                {
                    int t = board[i][j] - '0';
                    row[i][t] = col[j][t] = cell[i / 3][j / 3][t] = true;
                }
            }

        dfs(board,row,col,cell,0,0);
    }
}