36. Valid Sudoku

Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

Each row must contain the digits 1-9 without repetition.
Each column must contain the digits 1-9 without repetition.
Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition.

A partially filled sudoku which is valid.

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

Example 1:

Input:
[
  ["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:
[
  ["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.

Note:

1. A Sudoku board (partially filled) could be valid but is not necessarily solvable.
2. Only the filled cells need to be validated according to the mentioned rules.
3. The given board contain only digits 1-9 and the character ‘.’.
4. The given board size is always 9x9.

方法1: 3 次遍历

basketking：https://www.youtube.com/watch?v=iqe1JSjyldo

思路：

进行rowwise，colwise，cubewise的三次遍历，每次用一个set记住当前row/col/cube出现过哪些数字，一旦出现重复return false 就可以了。这个题的关键是怎么转换成cube的坐标：把i堪称在大九宫格中的循环，j在每一个小九宫格中的循环，他们全部需要按照 (num / 3, num % 3)的顺序完成。因此只需要对i ， j 取函数:
r = ( i / 3) * 3 + (j / 3)
c = (i % 3) * 3 + (j % 3)
即为转换到cubewise遍历的坐标。

易错点

1. 上面的转换：为什么要把 (i / 3)或者（i%3）再*3呢？因为每次i的变化是对于大九宫格，相当于向右或向下一次性前进3格。
2. set更新的位置
3. 要判断当前点是否是’.’：是的话跳过不作处理

Complexity

Time complexity: O(n^2)
Space complexity: O(1)

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {  
        for (int i = 0; i < 9; i++){
            set<int> row;
            for (int j = 0; j < 9; j ++){
                if (row.find(board[i][j]) != row.end()) return false;
                if (board[i][j] != '.')
                    row.insert(board[i][j]);
                
            }
        }
        
        
        for (int j = 0; j < 9; j++){
            set<int> col;
            for (int i = 0; i < 9; i ++){
                if (col.find(board[i][j]) != col.end()) return false;
                if (board[i][j] != '.')
                    col.insert(board[i][j]);
            }
        }
        
        for (int i = 0; i < 9; i++){
            set<int> cube;
            for (int j = 0; j < 9; j ++){
                int r = (i / 3 ) * 3 + j / 3 ;
                int c = (i % 3 ) * 3 + j % 3;
                if (cube.find(board[r][c]) != cube.end()) return false;
                if (board[r][c] != '.')
                    cube.insert(board[r][c]);
            }
        }
        return true;
    }
};

// i = 0, j = 0 , 1, 2, 3, 4, 5, 6, 7, 8
// (0, 0), (0, 1), (0, 2), (1, 0), (1, 1) , (1, 2), (2, 0), (2, 1), (2, 2)

// i = 1, j = 0 , 1, 2, 3, 4, 5, 6, 7, 8
// (0, 3), (0, 4), (0, 5), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)

// i = 2, j = 0 , 1, 2, 3, 4, 5, 6, 7, 8
// (0, 6), (0, 7), (0, 8), (1, 6), (1, 7), (1, 8), (2, 6), (2, 7), (2, 8)

// i = 3, j = 0 , 1, 2, 3, 4, 5, 6, 7, 8
// (3, 0), (3, 1), (3, 2), (4, 0), (4, 1) , (4, 2), (5, 0), (5, 1), (5, 2)

// i = 4, j = 0 , 1, 2, 3, 4, 5, 6, 7, 8
// (3, 3), (3, 4), (3, 5), (4, 3), (4, 4) , (4, 5), (5, 3), (5, 4), (5, 5)

// i = 5, j = 0 , 1, 2, 3, 4, 5, 6, 7, 8
// (3, 6), (3, 7), (3, 8), (4, 6), (4, 7) , (4, 8), (5, 6), (5, 7), (5, 8)

// i = 6, j = 0 , 1, 2, 3, 4, 5, 6, 7, 8
// (6, 0), (6, 1), (6, 2), (7, 0), (7, 1) , (7, 2), (8, 0), (8, 1), (8, 2)

// i = 7, j = 0 , 1, 2, 3, 4, 5, 6, 7, 8
// (6, 3), (6, 4), (6, 5)

方法2: 1次遍历

YRB: https://www.cnblogs.com/yrbbest/p/4436318.html

思路：

不用把i 和 j 当成行列坐标，可以用下面的函数转化一次性在三个set里面都check一次。

Complexity

Time complexity: O(n^2)
Space complexity: O(1)

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) { 
        for (int i = 0; i < 9; i++){
            set<int> row;
            set<int> col;
            set<int> cube;
            for (int j = 0; j < 9; j ++){
                // row
                if (board[i][j] != '.' && row.find(board[i][j]) != row.end()) return false;
                else row.insert(board[i][j]);
                
                // col
                if (board[j][i] != '.' && col.find(board[j][i]) != col.end()) return false;
                else col.insert(board[j][i]);
                
                // cube
                int r = (i / 3) * 3 + (j / 3);
                int c = (i % 3) * 3 + (j % 3);
                if (board[r][c] != '.' && cube.find(board[r][c]) != cube.end()) return false;
                else cube.insert(board[r][c]);
            }
        }
         
        return true;
    }
};