2021-06-01

37.解数独

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：

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：

提示：

board.length == 9
board[i].length == 9
board[i][j] 是一位数字或者 '.'
题目数据 保证 输入数独仅有一个解

 

题解1：递归+回溯

import java.util.ArrayList;
import java.util.List;

/*
 * @lc app=leetcode.cn id=37 lang=java
 *
 * [37] 解数独
 */

// @lc code=start
class Solution {
    boolean[][] row = new boolean[9][9];
    boolean[][] col = new boolean[9][9];
    boolean[][][] box = new boolean[3][3][9];
    List<int[]> spaces = new ArrayList<>();
    boolean isValid = false;

    public void solveSudoku(char[][] board) {
        // 深度优先遍历+回溯
        // 时间复杂度:O()
        // 空间复杂度:O()
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] == '.') {
                    spaces.add(new int[] { i, j });
                } else {
                    int digit = board[i][j] - '0' - 1;
                    row[i][digit] = true;
                    col[j][digit] = true;
                    box[i / 3][j / 3][digit] = true;
                }
            }
        }
        bfs(board, 0);
    }

    private void bfs(char[][] board, int pos) {
        if (pos == spaces.size()) {
            isValid = true;
            return;
        }
        int[] space = spaces.get(pos);
        int i = space[0], j = space[1];
        for (int digit = 0; digit < 9 && !isValid; digit++) {
            if (!row[i][digit] && !col[j][digit] && !box[i / 3][j / 3][digit]) {
                row[i][digit] = true;
                col[j][digit] = true;
                box[i / 3][j / 3][digit] = true;
                board[i][j] = (char) (digit + '0' + 1);
                bfs(board, pos + 1);
                row[i][digit] = false;
                col[j][digit] = false;
                box[i / 3][j / 3][digit] = false;
            }
        }
    }
}