Discuss：www.cnblogs.com/grandyang/p…
Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy all of the following rules:
- Each of the digits
1-9must occur exactly once in each row. - Each of the digits
1-9must occur exactly once in each column. - Each of the digits
1-9must occur exactly once in each of the 93x3sub-boxes of the grid.
The '.' character indicates empty cells.
Example 1:
Input: 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"]]
Output: [["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"]]
Explanation: The input board is shown above and the only valid solution is shown below:
Constraints:
board.length == 9board[i].length == 9board[i][j]is a digit or'.'.- It is guaranteed that the input board has only one solution.
解法一
这道求解数独的题是在之前那道 LeetCode 36. Valid Sudoku 验证数独 的基础上的延伸，之前那道题让我们验证给定的数组是否为数独数组，这道让求解数独数组。对于每个需要填数字的格子带入1到9，每代入一个数字都判定其是否合法，如果合法就继续下一次递归，结束时把数字设回 '.'，判断新加入的数字是否合法时，只需要判定当前数字是否合法，不需要判定这个数组是否为数独数组，因为之前加进的数字都是合法的，这样可以使程序更加高效一些，整体思路是这样的，但是实现起来可以有不同的形式。一种实现形式是递归带上横纵坐标，由于是一行一行的填数字，且是从0行开始的，所以当 i 到达 9 的时候，说明所有的数字都成功的填入了，直接返回 ture。当 j 大于等于 9 时，当前行填完了，需要换到下一行继续填，则继续调用递归函数，横坐标带入 i+1。否则看若当前数字不为点，说明当前位置不需要填数字，则对右边的位置调用递归。若当前位置需要填数字，则应该尝试填入1到9内的所有数字，让 c 从 1 遍历到 9，每当试着填入一个数字，都需要检验是否有冲突，使用另一个子函数 isValid 来检验是否合法，假如不合法，则跳过当前数字。若合法，则将当前位置赋值为这个数字，并对右边位置调用递归，若递归函数返回 true，则说明可以成功填充，直接返回 true。不行的话，需要重置状态，将当前位置恢复为点。若所有数字都尝试了，还是不行，则最终返回 false。参见代码如下：
class Solution {
fun solveSudoku(board: Array<CharArray>): Unit {
if (board.isEmpty()) {
return
}
solve(board)
}
private fun solve(
board: Array<CharArray>
): Boolean {
for (i in board.indices) {
for (j in board[0].indices) {
if (board[i][j] == '.') {
for (c in '1'..'9') {
if (isValid(board, i, j, c)) {
board[i][j] = c
if (solve(board)) {
return true
} else {
board[i][j] = '.'
}
}
}
return false
}
}
}
return true
}
private fun isValid(board: Array<CharArray>, row: Int, col: Int, c: Char): Boolean {
for (index in board.indices) {
if (board[index][col] != '.' && board[index][col] == c) {
return false
}
if (board[row][index] != '.' && board[row][index] == c) {
return false
}
if (board[3 * (row / 3) + index / 3][3 * (col / 3) + index % 3] != '.' &&
board[3 * (row / 3) + index / 3][3 * (col / 3) + index % 3] == c
) {
return false
}
}
return true
}
}
