LeetCode关于数独Sudoku的问题

关于我的 Leetcode 题目解答，代码前往 Github：https://github.com/chenxiangcyr/leetcode-answers

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LeetCode题目：36. Valid Sudoku
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 '.'.

Valid Sudoku
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.

class Solution {
    public boolean isValidSudoku(char[][] board) {
        int n = board.length;
        
        for(int i = 0; i < n; i++) {
            Set<Character> rows = new HashSet<Character>();
            Set<Character> columns = new HashSet<Character>();
            Set<Character> cubes = new HashSet<Character>();
            
            for(int j = 0; j < n; j++) {
                if(board[i][j] != '.') {
                    // check each row
                    if(rows.contains(board[i][j])) {
                        return false;
                    }
                    else {
                        rows.add(board[i][j]);
                    }
                }
                
                if(board[j][i] != '.') {
                    // check each column
                    if(columns.contains(board[j][i])) {
                        return false;
                    }
                    else {
                        columns.add(board[j][i]);
                    }
                }
                
                int rowIdx = 3 * (i / 3);
                int columnIdx = 3 * (i % 3);
                if(board[rowIdx + j / 3][columnIdx + j % 3] != '.') {
                    // check each cube
                    if(cubes.contains(board[rowIdx + j / 3][columnIdx + j % 3])) {
                        return false;
                    }
                    else {
                        cubes.add(board[rowIdx + j / 3][columnIdx + j % 3]);
                    }
                }
            }
        }
        
        return true;
    }
}

LeetCode题目：37. Sudoku Solver
Write a program to solve a Sudoku puzzle by filling the empty cells.
Empty cells are indicated by the character '.'.
You may assume that there will be only one unique solution.

Sudoku Solver Sudoku Solver

class Solution {
    public void solveSudoku(char[][] board) {
        if(board == null || board.length == 0 || board[0].length == 0) {
            return;
        }
        
        solve(board);
    }
    
    public boolean solve(char[][] board){
        for(int i = 0; i < board.length; i++) {
            for(int j = 0; j < board[0].length; j++) {
                if(board[i][j] == '.') {
                    // 尝试每一个数字
                    for(char c = '1'; c <= '9'; c++) {
                        if(isValid(board, i, j, c)){
                            board[i][j] = c;
                            
                            // 递归，处理下一个位置
                            if(solve(board)) {
                                return true;
                            }
                            else {
                                // 回溯 backtracking
                                board[i][j] = '.';
                            }
                        }
                    }
                    
                    return false;
                }
            }
        }
        
        return true;
    }
    
    // 把(row, col)位置设为c，判断是不是合法的数独
    private boolean isValid(char[][] board, int row, int col, char c){
        for(int i = 0; i < 9; i++) {
            // 检查第col列，不能有重复的c
            if(board[i][col] != '.' && board[i][col] == c) {
                return false;
            }
            
            // 检查第row行，不能有重复的c
            if(board[row][i] != '.' && board[row][i] == c) {
                return false;
            }
            
            // 检查对应的cube，不能有重复的c
            if(board[3 * (row / 3) + i / 3][ 3 * (col / 3) + i % 3] != '.' && 
board[3 * (row / 3) + i / 3][3 * (col / 3) + i % 3] == c) {
                return false;
            }
        }
        
        return true;
    }
}