Valid Sudoku

描述

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules [http://sudoku.com.au/TheRules.aspx]url(http://sudoku.com.au/TheRules.aspx) .

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

分析

细节实现题。

代码

Python

// Valid Sudoku
// 时间复杂度O(n^2)，空间复杂度O(1)
public class Solution {
    public boolean isValidSudoku(char[][] board) {
        boolean[] used = new boolean[9];

        for (int i = 0; i < 9; ++i) {
            Arrays.fill(used, false);

            for (int j = 0; j < 9; ++j) // 检查行
                if (!check(board[i][j], used))
                    return false;

            Arrays.fill(used, false);

            for (int j = 0; j < 9; ++j) // 检查列
                if (!check(board[j][i], used))
                    return false;
        }

        for (int r = 0; r < 3; ++r) // 检查 9 个子格子
            for (int c = 0; c < 3; ++c) {
                Arrays.fill(used, false);

                for (int i = r * 3; i < r * 3 + 3; ++i)
                    for (int j = c * 3; j < c * 3 + 3; ++j)
                        if (!check(board[i][j], used))
                            return false;
            }

        return true;
    }

    private static boolean check(char ch, boolean[] used) {
        if (ch == '.') return true;

        if (used[ch - '1']) return false;

        return used[ch - '1'] = true;
    }
};

Java

// Valid Sudoku
// 时间复杂度O(n^2)，空间复杂度O(1)
class Solution {
public:
    bool isValidSudoku(const vector<vector<char>>& board) {
        bool used[9];

        for (int i = 0; i < 9; ++i) {
            fill(used, used + 9, false);

            for (int j = 0; j < 9; ++j) // 检查行
                if (!check(board[i][j], used))
                    return false;

            fill(used, used + 9, false);

            for (int j = 0; j < 9; ++j) // 检查列
                if (!check(board[j][i], used))
                    return false;
        }

        for (int r = 0; r < 3; ++r) // 检查 9 个子格子
            for (int c = 0; c < 3; ++c) {
                fill(used, used + 9, false);

                for (int i = r * 3; i < r * 3 + 3; ++i)
                    for (int j = c * 3; j < c * 3 + 3; ++j)
                        if (!check(board[i][j], used))
                            return false;
            }

        return true;
    }

    bool check(char ch, bool used[9]) {
        if (ch == '.') return true;

        if (used[ch - '1']) return false;

        return used[ch - '1'] = true;
    }
};

C++

# Valid Sudoku
# Time complexity O(n^2), Space complexity O(1)
class Solution:
    def isValidSudoku(self, board: list[list[str]]) -> bool:
        used = [False] * 9

        for i in range(9):
            used = [False] * 9

            for j in range(9):  # check rows
                if not self.check(board[i][j], used):
                    return False

            used = [False] * 9

            for j in range(9):  # check columns
                if not self.check(board[j][i], used):
                    return False

        for r in range(3):  # check 9 sub-boxes
            for c in range(3):
                used = [False] * 9

                for i in range(r * 3, r * 3 + 3):
                    for j in range(c * 3, c * 3 + 3):
                        if not self.check(board[i][j], used):
                            return False

        return True

    @staticmethod
    def check(ch: str, used: list[bool]) -> bool:
        if ch == '.':
            return True

        idx = ord(ch) - ord('1')
        if used[idx]:
            return False

        used[idx] = True
        return True

相关题目

• Sudoku Solver