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37. Sudoku Solver

LeetCode - Sudoku Solver

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 1-9 must occur exactly once in each row.

  • Each of the digits 1-9 must occur exactly once in each column.

  • Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.

Empty cells are indicated by the character '.'.

0037 01

A sudoku puzzle…​

0037 02

…​and its solution numbers marked in red.

Note:

  • The given board contain only digits 1-9 and the character '.'.

  • You may assume that the given Sudoku puzzle will have a single unique solution.

  • The given board size is always 9x9.

解题分析

这道题特别需要关注的点是：只要找到一个解就可以了，不需要求全部解。所以，需要重点思考的是，如何在找到第一个解时，就立即停止搜索。

递归树

0037 03

  • 一刷

  • 二刷

  • 二刷（引入步进）

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/**
 *  * @author D瓜哥 · https://www.diguage.com
 *  * @since 2020-03-25 09:34
 */
public void solveSudoku(char[][] board) {
    backtrack(board, 0);
}

/**
 * Runtime: 17 ms, faster than 35.23% of Java online submissions for Sudoku Solver.
 * Memory Usage: 37 MB, less than 21.05% of Java online submissions for Sudoku Solver.
 */
private boolean backtrack(char[][] board, int step) {
    int len = board.length;
    if (step == len * len) {
        return true;
    }
    int y = step / len;
    int x = step % len;

    for (int i = y; i < len; i++) {
        for (int j = x; j < len; j++) {
            if (board[i][j] != '.') {
                return backtrack(board, step + 1);
            }
            for (char c = '1'; c <= '9'; c++) {
                if (!isValid(board, i, j, c)) {
                    continue;
                }
                board[i][j] = c;
                System.out.printf("y=%d,x=%d, num=%s, step=%d %n",
                        y, x, c, step);
                Printers.printMatrix(board);
                if (backtrack(board, step + 1)) {
                    return true;
                }
                board[i][j] = '.';
            }
            return false;
        }
    }
    return false;
}

private boolean isValid(char[][] board, int y, int x, char c) {
    for (int i = 0; i < 9; i++) {
        if (board[(y / 3) * 3 + i / 3][(x / 3) * 3 + i % 3] == c
                || board[y][i] == c
                || board[i][x] == c) {
            return false;
        }
    }
    return true;
}

/**
 * Copy From
 */
private boolean backtrackXY(char[][] board, int y, int x) {
    int len = board.length;
    if (x == len) {
        return backtrackXY(board, y + 1, 0);
    }
    if (y == len) {
        return true;
    }
    for (int i = y; i < len; i++) {
        for (int j = x; j < len; j++) {
            if (board[i][j] != '.') {
                return backtrackXY(board, i, j + 1);
            }
            for (char c = '1'; c <= '9'; c++) {
                if (!isValid(board, i, j, c)) {
                    continue;
                }
                board[i][j] = c;
                System.out.printf("y=%d,x=%d, num=%s %n", i, j, c);
                Printers.printMatrix(board);
                if (backtrackXY(board, i, j + 1)) {
                    return true;
                }
                board[i][j] = '.';
            }
            return false;
        }
    }
    return false;
}

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/**
 * * @author D瓜哥 · https://www.diguage.com
 * * @since 2024-09-11 10:59:08
 */
public void solveSudoku(char[][] board) {
  backtracking(board);
}

private boolean backtracking(char[][] board) {
  for (int r = 0; r < board.length; r++) {
    for (int c = 0; c < board[r].length; c++) {
      if (board[r][c] != '.') {
        continue;
      }
      // 根据限制条件，筛选出可选字符，否则重复判断
      char[] chars = selectChars(board, r, c);
      for (char sc : chars) {
        board[r][c] = sc;
        printMatrix(board);
        if (backtracking(board)) {
          return true;
        }
        board[r][c] = '.';
      }
      return false;
    }
  }
  return true;
}

private char[] selectChars(char[][] board, int row, int column) {
  char[] chars = new char[9];
  int length = 9;
  for (int i = 0; i < chars.length; i++) {
    chars[i] = (char) ('1' + i);
  }
  for (int i = 0; i < board.length; i++) {
    char c = board[i][column];
    if (c != '.' && chars[c - '1'] != '.') {
      chars[c - '1'] = '.';
      length--;
      if (length == 0) {
        return new char[0];
      }
    }
  }
  for (int i = 0; i < board[row].length; i++) {
    char c = board[row][i];
    if (c != '.' && chars[c - '1'] != '.') {
      chars[c - '1'] = '.';
      length--;
      if (length == 0) {
        return new char[0];
      }
    }
  }
  row = (row / 3) * 3;
  column = (column / 3) * 3;
  for (int i = row; i < row + 3; i++) {
    for (int j = column; j < column + 3; j++) {
      char c = board[i][j];
      if (c != '.' && chars[c - '1'] != '.') {
        chars[c - '1'] = '.';
        length--;
        if (length == 0) {
          return new char[0];
        }
      }
    }
  }
  char[] result = new char[length];
  for (int i = chars.length - 1; i >= 0; i--) {
    if (chars[i] != '.') {
      result[--length] = chars[i];
    }
  }
  return result;
}

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/**
 * * @author D瓜哥 · https://www.diguage.com
 * * @since 2024-09-11 10:59:08
 */
public void solveSudoku(char[][] board) {
  backtracking(board, 0);
}

private boolean backtracking(char[][] board, int step) {
  int row = step / board.length;
  int column = step % board.length;
  for (int r = row; r < board.length; r++) {
    for (int c = column; c < board[r].length; c++) {
      if (board[r][c] != '.') {
        return backtracking(board, step + 1);
      }
      // 根据限制条件，筛选出可选字符，否则重复判断
      char[] chars = selectChars(board, r, c);
      for (char sc : chars) {
        board[r][c] = sc;
        printMatrix(board);
        if (backtracking(board, step + 1)) {
          return true;
        }
        board[r][c] = '.';
      }
      return false;
    }
  }
  return true;
}

private char[] selectChars(char[][] board, int row, int column) {
  char[] chars = new char[9];
  int length = 9;
  for (int i = 0; i < chars.length; i++) {
    chars[i] = (char) ('1' + i);
  }
  for (int i = 0; i < board.length; i++) {
    char c = board[i][column];
    if (c != '.' && chars[c - '1'] != '.') {
      chars[c - '1'] = '.';
      length--;
      if (length == 0) {
        return new char[0];
      }
    }
  }
  for (int i = 0; i < board[row].length; i++) {
    char c = board[row][i];
    if (c != '.' && chars[c - '1'] != '.') {
      chars[c - '1'] = '.';
      length--;
      if (length == 0) {
        return new char[0];
      }
    }
  }
  row = (row / 3) * 3;
  column = (column / 3) * 3;
  for (int i = row; i < row + 3; i++) {
    for (int j = column; j < column + 3; j++) {
      char c = board[i][j];
      if (c != '.' && chars[c - '1'] != '.') {
        chars[c - '1'] = '.';
        length--;
        if (length == 0) {
          return new char[0];
        }
      }
    }
  }
  char[] result = new char[length];
  for (int i = chars.length - 1; i >= 0; i--) {
    if (chars[i] != '.') {
      result[--length] = chars[i];
    }
  }
  return result;
}

参考资料

  1. 搞懂回溯算法，我终于能做数独了

  2. 37. 解数独 - 「代码随想录」带你学透回溯算法！

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