Leetcode: Valid Sudoku

最新推荐文章于 2025-08-12 12:49:47 发布

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最新推荐文章于 2025-08-12 12:49:47 发布

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分类专栏： Leetcode 文章标签： leetcode

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本文介绍了一种检查数独是否有效的算法实现。该算法通过检查数独的每一行、每一列以及每一个九宫格内数字是否重复来判断数独是否有效。文章提供了完整的Java代码示例。

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Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

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

A partially filled sudoku which is valid.

Note:

A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

Use the three rules to check for each row, column, and block. Need O(3*9^2) time.

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        if (board == null || board.length != 9 || board[0].length != 9) {
            return false;
        }
        
        for (int i = 0; i < 9; i++) {
            boolean[] map = new boolean[9];
            for (int j = 0; j < 9;j++) {
                if (board[i][j] != '.') {
                    if (map[board[i][j] - '1']) {
                        return false;
                    }
                    map[board[i][j] - '1'] = true;
                }
            }
        }
        
        for (int i = 0; i < 9; i++) {
            boolean[] map = new boolean[9];
            for (int j = 0; j < 9; j++) {
                if (board[j][i] != '.') {
                    if (map[board[j][i] - '1']) {
                        return false;
                    }
                    map[board[j][i] - '1'] = true;
                }
            }
        }
        
        for (int i = 0; i < 9; i++) {
            boolean[] map = new boolean[9];
            for (int j = 0; j < 9; j++) {
                char curr = board[i / 3 * 3 + j / 3][i % 3 * 3 + j % 3];
                if (curr != '.') {
                    if (map[curr - '1']) {
                        return false;
                    }
                    map[curr - '1'] = true;
                }
            }
        }
        
        return true;
    }
}