leetcode: Valid Sudoku

http://oj.leetcode.com/problems/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 '.'.

A partially filled sudoku which is valid.

思路：

按行，按列，再按3 * 3小方块检查，只要都通过就返回true，不然就是false。

class Solution {
public:
    bool isValidSudoku(vector<vector<char> > &board) {
        int row, col, block;
        char ch;
        vector<int> marks(9, 0);
        
        // Rows
        for (row = 0; row < 9; ++row) {
            marks.assign(9, 0);
            
            for (col = 0; col < 9; ++col) {
                ch = board[row][col];
                
                if (ch != '.') {
                    if (marks[ch - '1'] > 0) {
                        return false;
                    }
                    else {
                        ++marks[ch - '1'];
                    }
                }
            }
        }
        
        // Cols
        for (col = 0; col < 9; ++col) {
            marks.assign(9, 0);
            
            for (row = 0; row < 9; ++row) {
                ch = board[row][col];
                
                if (ch != '.') {
                    if (marks[ch - '1'] > 0) {
                        return false;
                    }
                    else {
                        ++marks[ch - '1'];
                    }
                }
            }
        }
        
        // Blocks
        for (block = 0; block < 9; ++block) {
            int start_row = (block / 3) * 3, start_col = (block % 3) * 3;
            marks.assign(9, 0);
            
            for (row = start_row; row < (start_row + 3); ++row) {
                for (col = start_col; col < (start_col + 3); ++col) {
                    ch = board[row][col];

                    if (ch != '.') {
                        if (marks[ch - '1'] > 0) {
                            return false;
                        }
                        else {
                            ++marks[ch - '1'];
                        }
                    }
                }
            }
        }
        
        return true;
    }
};