LeetCode(36)-Valid Sudoku

问题描述：

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.

问题分析：

判断数独是否有效，不要求求解。

创建一个二维数组存储，只要同一行、列和块没有元素没有相同的元素即可。

问题求解：

   bool isValidSudoku(vector<vector<char>>& board) {    vector<short> col(9, 0);    vector<short> block(9, 0);    vector<short> row(9, 0);    for (int i = 0; i < 9; i++)     for (int j = 0; j < 9; j++) {         if (board[i][j] != '.') {             int idx = 1 << (board[i][j] - '0');             if (row[i] & idx || col[j] & idx || block[i/3 * 3 + j / 3] & idx)                return false;            row[i] |= idx;            col[j] |= idx;            block[i/3 * 3 + j/3] |= idx;         }     }     return true;  }

第二遍：

class Solution {public:    bool isValidSudoku(vector<vector<char>>& board) {    vector<unordered_set<char>> rows(9), cols(9), grid(9);    for (int i = 0; i < 9; i++){     for (int j = 0; j < 9; j++) {         if (board[i][j] != '.') {            if(rows[i].find(board[i][j]) != rows[i].end())            {                return false;            }else{                rows[i].insert(board[i][j]);            }            if(cols[j].find(board[i][j]) != rows[j].end())            {                return false;            }else{                cols[j].insert(board[i][j]);            }                        int n = i-i%3+j/3;           if(grid[n].find(board[i][j]) != grid[n].end())            {                return false;            }else{                grid[n].insert(board[i][j]);            }         }     }    }     return true;    }};

这次用hash set求解，关键是行列和网格要选合适的数据结构表示，还有n= i-i%3+j/3。从该题了解了我对STL掌握不是很好，接下来要好好总结。