[LeetCode] 36. Valid Sudoku

Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

1. Each row must contain the digits 1-9 without repetition.
2. Each column must contain the digits 1-9 without repetition.
3. Each of the nine 3 x 3 sub-boxes of the grid must contain the digits 1-9 without repetition.

Note:

• A Sudoku board (partially filled) could be valid but is not necessarily solvable.
• Only the filled cells need to be validated according to the mentioned rules.

Example 1:

Input: 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"]]
Output: true

Example 2:

Input: board = 
[["8","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"]]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Constraints:

• board.length == 9
• board[i].length == 9
• board[i][j] is a digit or '.'.

有效的数独。

题意如下

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

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。

这个题没有什么算法可言，思路是用hashmap判断每一行，每一列，和每一个九宫格是否满足条件。如果当前的坐标内不是点（.），且当前坐标内的数字存在于hashmap则return false。判断每一行的时候，因为坐标是（i, j），所以判断（j, i）的时候可以同时判断一个对应的列是否满足题意。判断每个cube的方式是先要得到rowIndex和colIndex，这里我是当做结论一样背下来的，真的不是很好记。

• rowIndex = 3 * (i / 3)
• colIndex = 3 * (i % 3)

时间O(mn)

空间O(n)

Java实现

class Solution {
    public boolean isValidSudoku(char[][] board) {
        for (int i = 0; i < board.length; i++) {
            HashSet<Character> rows = new HashSet<>();
            HashSet<Character> cols = new HashSet<>();
            HashSet<Character> cube = new HashSet<>();
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] != '.' && !rows.add(board[i][j])) {
                    return false;
                }
                if (board[j][i] != '.' && !cols.add(board[j][i])) {
                    return false;
                }
                int rowIndex = 3 * (i / 3);
                int colIndex = 3 * (i % 3);
                if (board[rowIndex + j / 3][colIndex + j % 3] != '.'
                        && !cube.add(board[rowIndex + j / 3][colIndex + j % 3])) {
                    return false;
                }
            }
        }
        return true;
    }
}

JavaScript实现

/**
 * @param {character[][]} board
 * @return {boolean}
 */
var isValidSudoku = function(board) {
    let m = board.length;
    let n = board[0].length;
    for (let i = 0; i < m; i++) {
        let rows = new Set();
        let cols = new Set();
        let cubes = new Set();
        for (let j = 0; j < n; j++) {
            if (board[i][j] != '.') {
                if (!rows.has(board[i][j])) {
                    rows.add(board[i][j]);
                } else {
                    return false;
                }
            }
            if (board[j][i] != '.') {
                if (!cols.has(board[j][i])) {
                    cols.add(board[j][i]);
                } else {
                    return false;
                }
            }
            let rowIndex = 3 * Math.floor(i / 3);
            let colIndex = 3 * (i % 3);
            if (
                board[rowIndex + Math.floor(j / 3)][colIndex + (j % 3)] != '.'
            ) {
                if (
                    !cubes.has(
                        board[rowIndex + Math.floor(j / 3)][colIndex + (j % 3)]
                    )
                ) {
                    cubes.add(
                        board[rowIndex + Math.floor(j / 3)][colIndex + (j % 3)]
                    );
                } else {
                    return false;
                }
            }
        }
    }
    return true;
};

2021年4月更新

我这里再推荐一个字符串的思路，来自于StefanPochmann大神。这个做法的思想是需要判断某个元素是否是 unique 的，这个思想可以运用到其他需要判断独一性的题目。这里我一时想不起来有哪些题目了，日后有总结我再加以更新。

Java实现

class Solution {
    public boolean isValidSudoku(char[][] board) {
        HashSet<String> set = new HashSet<>();
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] != '.') {
                    String b = "(" + board[i][j] + ")";
                    if (!set.add(b + i) || !set.add(j + b) || !set.add(i / 3 + b + j / 3)) {
                        return false;
                    }
                }
            }
        }
        return true;
    }
}

LeetCode 题目总结