问题

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

示例：
leetcode-37：解数独 - 图1
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"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：
leetcode-37：解数独 - 图2

思路

棋盘搜索问题可以使用回溯法暴力搜索，只不过这次我们要做的是二维递归，怎么做二维递归呢？

N皇后问题是因为每一行每一列只放一个皇后，只需要一层for循环遍历一行，递归来来遍历列，然后一行一列确定皇后的唯一位置

本题就不一样了，本题中棋盘的每一个位置都要放一个数字，并检查数字是否合法，解数独的树形结构要比N皇后更宽更深

因为这个树形结构太大了，抽取一部分，如图所示：
leetcode-37：解数独 - 图3

回溯三部曲

  • 递归函数以及参数

    • 递归函数的返回值需要是**bool**类型，为什么呢？
      • 因为解数独找到一个符合的条件（就在树的叶子节点上）立刻就返回，相当于找从根节点到叶子节点一条唯一路径，所以需要使用bool返回值
        1. public boolean backtracking(char[][] board)

  • 递归终止条件

    • 本题递归不用终止条件，解数独是要遍历整个树形结构寻找可能的叶子节点就立刻返回
    • 不用终止条件会不会死循环
      • 递归的下一层的棋盘一定比上一层的棋盘多一个数，等数填满了棋盘自然就终止（填满当然好了，说明找到结果了），所以不需要终止条件

  • 递归单层搜索逻辑

leetcode-37：解数独 - 图4
在树形图中可以看出我们需要的是一个二维递归（也就是两个for循环嵌套着递归）

一个**for**循环遍历棋盘的行，一个**for**循环遍历棋盘的列，一行一列确定下来之后，递归遍历这个位置放9个数字的可能性

  1. public boolean backtracking(char[][] board) {
  2.     for (int i = 0; i < board.length; i++) {        // 遍历行
  3.         for (int j = 0; j < board[0].length; j++) { // 遍历列
  4.             if (board[i][j] != '.') 
  5.                 continue;
  6.             for (char k = '1'; k <= '9'; k++) {     // (i, j) 这个位置放k是否合适
  7.                 if (isValid(i, j, k, board)) { 
  8.                     board[i][j] = k;                // 放置k
  9.                     if (backtracking(board)) 
  10.                         return true; // 如果找到合适一组立刻返回
  11.                     board[i][j] = '.';              // 回溯，撤销k
  12.                 }
  13.             }
  14.             return false; // 9个数都试完了，都不行，那么就返回false
  15.         }
  16.     }
  17.     return true; // 遍历完没有返回false，说明找到了合适棋盘位置了
  18. }
  • 如一定义二维数组array[][],则获得该数组的长度（即行数）的代码为array.length
  • 对于每一行的宽度可以循环获得，如获得第i行的宽度(i的范围是0~array.length-1)的代码为array[i].length

注意这里return false的地方，这里放return false 是有讲究的

  • 因为如果一行一列确定下来了，这里尝试了9个数都不行，说明这个棋盘找不到解决数独问题的解！那么会直接返回，这也就是为什么没有终止条件也不会永远填不满棋盘而无限递归下去

判断棋盘是否合法

判断棋盘是否合法有如下三个维度：

  • 同行是否重复
  • 同列是否重复

  • 9宫格里是否重复

    1. public boolean isValid(int row, int col, int k, char[][] board) {
    2.   for (int i = 0; i < 9; i++) { // 判断行里是否重复
    3.       if (board[row][i] == k) {
    4.           return false;
    5.       }
    6.   }
    7.   for (int j = 0; j < 9; j++) { // 判断列里是否重复
    8.       if (board[j][col] == k) {
    9.           return false;
    10.       }
    11.   }
    13.   int startRow = (row / 3) * 3;
    14.   int startCol = (col / 3) * 3;
    15.   for (int i = startRow; i < startRow + 3; i++) { // 判断9方格里是否重复
    16.       for (int j = startCol; j < startCol + 3; j++) {
    17.           if (board[i][j] == val ) {
    18.               return false;
    19.           }
    20.       }
    21.   }
    22.   return true;
    23. }

    ```java class Solution { public boolean backtracking(char[][] board) { for (int i = 0; i < board.length; i++) { // 遍历行

    1.   for (int j = 0; j < board[0].length; j++) { // 遍历列
    2.       if (board[i][j] != '.') 
    3.           continue;
    4.       for (char k = '1'; k <= '9'; k++) {     // (i, j) 这个位置放k是否合适
    5.           if (isValid(i, j, k, board)) { 
    6.               board[i][j] = k;                // 放置k
    7.               if (backtracking(board)) 
    8.                   return true; // 如果找到合适一组立刻返回
    9.               board[i][j] = '.';              // 回溯，撤销k
    10.           }
    11.       }
    12.       return false; // 9个数都试完了，都不行，那么就返回false
    13.   }

    } return true; // 遍历完没有返回false，说明找到了合适棋盘位置了 }

public boolean isValid(int row, int col, int k, char[][] board) { for (int i = 0; i < 9; i++) { // 判断行里是否重复 if (board[row][i] == k) { return false; } } for (int j = 0; j < 9; j++) { // 判断列里是否重复 if (board[j][col] == k) { return false; } }

int startRow = (row / 3) * 3;
int startCol = (col / 3) * 3;
for (int i = startRow; i < startRow + 3; i++) { // 判断9方格里是否重复
    for (int j = startCol; j < startCol + 3; j++) {
        if (board[i][j] == k ) {
            return false;
        }
    }
}
return true;

}

public void solveSudoku(char[][] board) {
    backtracking(board);
}

} ```