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37. 解数独

English Version

题目描述

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：

  1. 数字 1-9 在每一行只能出现一次。
  2. 数字 1-9 在每一列只能出现一次。
  3. 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 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：

 

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

解法

方法一：回溯

我们用数组 row、col、box 分别记录每一行、每一列、每个 3x3 宫格中数字是否出现过。如果数字 i 在第 r 行、第 c 列、第 b 个 3x3 宫格中出现过，那么 row[r][i]、col[c][i]、box[b][i] 都为 true。

我们遍历 board 的每一个空格，枚举它可以填入的数字 v，如果 v 在当前行、当前列、当前 3x3 宫格中没有出现过，那么我们就可以尝试填入数字 v，并继续搜索下一个空格。如果搜索到最后，所有空格填充完毕，那么就说明找到了一个可行解。

时间复杂度 $O(9^{81})$，空间复杂度 $O(9^2)$。

Python3

class Solution:
    def solveSudoku(self, board: List[List[str]]) -> None:
        def dfs(k):
            nonlocal ok
            if k == len(t):
                ok = True
                return
            i, j = t[k]
            for v in range(9):
                if row[i][v] == col[j][v] == block[i // 3][j // 3][v] == False:
                    row[i][v] = col[j][v] = block[i // 3][j // 3][v] = True
                    board[i][j] = str(v + 1)
                    dfs(k + 1)
                    row[i][v] = col[j][v] = block[i // 3][j // 3][v] = False
                if ok:
                    return

        row = [[False] * 9 for _ in range(9)]
        col = [[False] * 9 for _ in range(9)]
        block = [[[False] * 9 for _ in range(3)] for _ in range(3)]
        t = []
        ok = False
        for i in range(9):
            for j in range(9):
                if board[i][j] == '.':
                    t.append((i, j))
                else:
                    v = int(board[i][j]) - 1
                    row[i][v] = col[j][v] = block[i // 3][j // 3][v] = True
        dfs(0)

Java

class Solution {
    private boolean ok;
    private char[][] board;
    private List<Integer> t = new ArrayList<>();
    private boolean[][] row = new boolean[9][9];
    private boolean[][] col = new boolean[9][9];
    private boolean[][][] block = new boolean[3][3][9];

    public void solveSudoku(char[][] board) {
        this.board = board;
        for (int i = 0; i < 9; ++i) {
            for (int j = 0; j < 9; ++j) {
                if (board[i][j] == '.') {
                    t.add(i * 9 + j);
                } else {
                    int v = board[i][j] - '1';
                    row[i][v] = col[j][v] = block[i / 3][j / 3][v] = true;
                }
            }
        }
        dfs(0);
    }

    private void dfs(int k) {
        if (k == t.size()) {
            ok = true;
            return;
        }
        int i = t.get(k) / 9, j = t.get(k) % 9;
        for (int v = 0; v < 9; ++v) {
            if (!row[i][v] && !col[j][v] && !block[i / 3][j / 3][v]) {
                row[i][v] = col[j][v] = block[i / 3][j / 3][v] = true;
                board[i][j] = (char) (v + '1');
                dfs(k + 1);
                row[i][v] = col[j][v] = block[i / 3][j / 3][v] = false;
            }
            if (ok) {
                return;
            }
        }
    }
}

C++

using pii = pair<int, int>;

class Solution {
public:
    void solveSudoku(vector<vector<char>>& board) {
        bool row[9][9] = {false};
        bool col[9][9] = {false};
        bool block[3][3][9] = {false};
        bool ok = false;
        vector<pii> t;
        for (int i = 0; i < 9; ++i) {
            for (int j = 0; j < 9; ++j) {
                if (board[i][j] == '.') {
                    t.push_back({i, j});
                } else {
                    int v = board[i][j] - '1';
                    row[i][v] = col[j][v] = block[i / 3][j / 3][v] = true;
                }
            }
        }
        function<void(int k)> dfs = [&](int k) {
            if (k == t.size()) {
                ok = true;
                return;
            }
            int i = t[k].first, j = t[k].second;
            for (int v = 0; v < 9; ++v) {
                if (!row[i][v] && !col[j][v] && !block[i / 3][j / 3][v]) {
                    row[i][v] = col[j][v] = block[i / 3][j / 3][v] = true;
                    board[i][j] = v + '1';
                    dfs(k + 1);
                    row[i][v] = col[j][v] = block[i / 3][j / 3][v] = false;
                }
                if (ok) {
                    return;
                }
            }
        };
        dfs(0);
    }
};

Go

func solveSudoku(board [][]byte) {
	var row, col [9][9]bool
	var block [3][3][9]bool
	var t [][2]int
	ok := false
	for i := 0; i < 9; i++ {
		for j := 0; j < 9; j++ {
			if board[i][j] == '.' {
				t = append(t, [2]int{i, j})
			} else {
				v := int(board[i][j] - '1')
				row[i][v], col[j][v], block[i/3][j/3][v] = true, true, true
			}
		}
	}
	var dfs func(int)
	dfs = func(k int) {
		if k == len(t) {
			ok = true
			return
		}
		i, j := t[k][0], t[k][1]
		for v := 0; v < 9; v++ {
			if !row[i][v] && !col[j][v] && !block[i/3][j/3][v] {
				row[i][v], col[j][v], block[i/3][j/3][v] = true, true, true
				board[i][j] = byte(v + '1')
				dfs(k + 1)
				row[i][v], col[j][v], block[i/3][j/3][v] = false, false, false
			}
			if ok {
				return
			}
		}
	}
	dfs(0)
}

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