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37. Sudoku Solver ​

题目 ​

Write a program to solve a Sudoku puzzle by filling the empty cells.

A sudoku solution must satisfy all of the following rules:

  1. Each of the digits 1-9 must occur exactly once in each row.
  2. Each of the digits 1-9 must occur exactly once in each column.
  3. Each of the digits 1-9 must occur exactly once in each of the 9 3x3 sub-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 == 9
  • board[i].length == 9
  • board[i][j] is a digit or '.'.
  • It is guaranteed that the input board has only one solution.
Related Topics
  • 数组
  • 哈希表
  • 回溯
  • 矩阵

  • 👍 1944
  • 👎 0

    • 思路 ​

    N 皇后问题的扩展，需要先遍历棋盘获取已知条件

    • 已存放的数字，空置的位置
    • 将存放数字存入对应的 Col Row 哈希表中
    • 空置位置存入到 blank 数组中
    • 进行回溯操作

    注意棋盘上的数字是 char 类型，不是 int 类型

    解法 ​

    python
    class Solution:
    def solveSudoku(self, board: List[List[str]]) -> None:
        GRID = 9
        col_map = {i: set() for i in range(GRID)}
        row_map = {i: set() for i in range(GRID)}
        box_map = {i: set() for i in range(GRID)}
        blank = []

        for row in range(GRID):
            for col in range(GRID):
                if board[row][col] != '.':
                    b = (row // 3) * 3 + (col // 3)
                    col_map[col].add(board[row][col])
                    row_map[row].add(board[row][col])
                    box_map[b].add(board[row][col])
                else:
                    blank.append([row, col])

        def backtrack(start):
            if start == len(blank):
                return True

            row, col = blank[start]
            box_index = (row // 3) * 3 + (col // 3)

            for ch in map(str, range(1, 10)):
                if ch in row_map[row] or ch in col_map[col] or ch in box_map[box_index]:
                    continue
                board[row][col] = ch
                col_map[col].add(board[row][col])
                row_map[row].add(board[row][col])
                box_map[box_index].add(board[row][col])
                if backtrack(start + 1):
                    return True
                col_map[col].remove(board[row][col])
                row_map[row].remove(board[row][col])
                box_map[box_index].remove(board[row][col])
                board[row][col] = '.'
            return False

        backtrack(0)

    复杂度分析 ​

    • 时间复杂度 O(9N)
      • N 为棋盘空位
    • 空间复杂度 O(N)