[Leetcode] valid Sudoku

Determine if a Sudoku is valid, according to: Sudoku puzzles-the rules.

The sudoku board cocould 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.

 

This question seems very complicated, but it is actually very brainless, because there is no need to judge whether it is a solvable Sudoku board. When this board is traversed and a position not empty is found, the rows, columns, and 3x3 are checked respectively. If a duplicate is displayed, the result is invalid. If the traversal is complete, the result is legal.

    bool isValidSudoku(vector<vector<char> > &board) {            unordered_set<int> column;    unordered_set<int> square;    unordered_set<int> row;         // check for colum    for (int i = 0; i < 9; i++) {        for (int j = 0; j < 9; j++) {            char item = board[j][i];            if (item != ‘.‘) {                int num = atoi(&item);                if (column.find(num) == column.end()){                    column.insert(atoi(&item));                }                else {                    return false;                }            }        }        column.clear();    }    // check for row    for (int i = 0; i < 9; i++) {        for (int j = 0; j < 9; j++) {            char item = board[i][j];            if (item != ‘.‘) {                int num = atoi(&item);                if (row.find(num) == row.end()) {                    row.insert(num);                }                else {                    return false;                }            }        }        row.clear();    }        // check for 3x3    for (int offsetY = 0; offsetY < 3; offsetY++) {        for (int offsetX = 0; offsetX < 3; offsetX++) {                        for (int i = 3*offsetY; i < 3*offsetY+3; i++) {                for (int j = 3*offsetX; j < 3*offsetX+3; j++) {                    char item = board[i][j];                    if (item != ‘.‘) {                        int num = atoi(&item);                        if (square.find(num) == square.end()) {                            square.insert(num);                        }                        else {                            return false;                        }                    }                }            }            square.clear();                    }    }        return true;    }

 

[Leetcode] valid Sudoku