36. Valid Sudoku

题目：

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.

链接： http://leetcode.com/problems/valid-sudoku/

题解：

验证数独，分三次验证行，列以及9个3x3子块就可以了。

Time Complexity - O(n²)， Space Complexity - O(1)。

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        if(board == null || board.length == 0)
            return false;
        HashSet<Character> set = new HashSet<Character>();
        
        for(int i = 0; i < 9; i++) {
            for(int j = 0; j < 9; j++) {
                if(board[i][j] != '.' && set.contains(board[i][j]))
                    return false;
                set.add(board[i][j]);
            }
            set.clear();
        }
        
        for(int j = 0; j < 9; j++) {
            for(int i = 0; i < 9; i++) {
                if(board[i][j] != '.' && set.contains(board[i][j]))
                    return false;
                set.add(board[i][j]);
            }
            set.clear();
        }
        
        for(int i = 1; i < 9; i += 3) {
            for(int j = 1; j < 9; j += 3) {
                for(int k = -1; k <= 1; k++) {
                    for(int l = -1; l <= 1; l++) {
                        if(board[i + k][j + l] != '.' && set.contains(board[i + k][j + l]))
                            return false;
                        set.add(board[i + k][j + l]);
                    }
                }
                set.clear();
            }
        }
         
        return true;   
    }
}

二刷：

Java:

依然是 Time Complexity - O(n²) ， Space Complexity - O(1)。这样遍历了三次数组，并不是很快，可以继续优化，只遍历一次数组。

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        if (board == null || board.length == 0) {
            return false;
        }
        Set<Character> set = new HashSet<>();
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] != '.' && !set.add(board[i][j])) {
                    return false;
                }
            }
            set.clear();
        }
        
        for (int j = 0; j < board[0].length; j++) {
            for (int i = 0; i < board.length; i++) {
                if (board[i][j] != '.' && !set.add(board[i][j])) {
                    return false;
                }
            }
            set.clear();
        }       
        
        for (int i = 1; i < board.length; i += 3) {
            for (int j = 1; j < board[0].length; j += 3) {
                for (int k = -1; k <= 1; k++) {
                    for (int l = -1; l <= 1; l++) {
                        if (board[i + k][j + l] != '.' && !set.add(board[i + k][j + l])) {
                            return false;
                        }
                    }
                }
                set.clear();
            }
        }
        return true;
    }
}

只遍历一次数组，代码主要来自于Lorraine921:

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        if (board == null || board.length == 0) {
            return false;
        }
        Set<Character> rows = new HashSet<Character>();
        Set<Character> cols = new HashSet<Character>();
        Set<Character> cube = new HashSet<Character>();
        for(int i = 0; i < 9; i++){
            for (int j = 0; j < 9;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;
                }
            }
            rows.clear();
            cols.clear();
            cube.clear();
        }
        return true;
    }
}

三刷:

看来三刷并没有吸取到二刷的精华，上面的解法完全忘记了。需要再领会。

Java:

public class Solution {
    public boolean isValidSudoku(char[][] board) {  // assume len = 9
        if (board == null || board.length == 0) {
            return false;
        }
        Set<Integer> set = new HashSet<>();
        for (int i = 0; i < 9; i++) {
            set.clear();
            for (int j = 0; j < 9; j++) {
                if (board[i][j] != '.' && !set.add(board[i][j] - '0')) {
                    return false;
                }
            }
        }
        for (int j = 0; j < 9; j++) {
            set.clear();
            for (int i = 0; i < 9; i++) {
                if (board[i][j] != '.' && !set.add(board[i][j] - '0')) {
                    return false;
                }
            }
        }
        for (int i = 1; i < 9; i += 3) {
            for (int j = 1; j < 9; j += 3) {
                set.clear();
                for (int k = -1; k <= 1; k++) {
                    for (int l = -1; l <= 1; l++) {
                        if (board[i + k][j + l] != '.' && !set.add(board[i + k][j + l] - '0')) {
                            return false;
                        }
                    }
                }
            }
        }
        return true;
    }
}

Reference:

https://leetcode.com/discuss/48737/1-7-lines-python-4-solutions

https://leetcode.com/discuss/27272/shared-my-concise-java-code

https://leetcode.com/discuss/23901/my-short-solution-by-c-o-n2

https://leetcode.com/discuss/17990/sharing-my-easy-understand-java-solution-using-set

https://leetcode.com/discuss/92445/yet-another-java-2ms-solution