[Leetcode]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.

Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

这是一道用到hash map的题目。

拿到这道题一上来的思路就是遍历所有元素，将每个元素都和同行、同列、同九宫格的元素比较，然后发现有很多重复操作。

然后想到可以用hash map（思路来源于Single Number II），将每行、每列、每九宫格出现的元素全部作为key存放在一个map中，如果当前行、列、方九宫格再次出现同一元素，则证明数独非法。

第一遍ac的代码（遍历三遍数组，分别确定行、列、九宫格是否合法）

class Solution {public:bool isValidSudoku(vector<vector<char> > &board) {int n = board.size();if(!rowValid(board, n)) return false;if(!colValid(board, n)) return false;if(!blockValid(board,n)) return false;return true;}bool rowValid(vector<vector<char> > board,int n){map<char, int> m;for (int i = 0; i < n; i++){for (int j = 0; j < n; j++){if (board[i][j] == '.') continue;if (m.find(board[i][j]) == m.end())m[board[i][j]] = 1;else return false;}m.clear();}return true;}bool colValid(vector<vector<char> > board, int n){map<char, int> m;for (int i = 0; i < n; i++){for (int j = 0; j < n; j++){if (board[j][i] == '.') continue;if (m.find(board[j][i]) == m.end())m[board[j][i]] = 1;else return false;}m.clear();}return true;}bool blockValid(vector<vector<char> > board, int n){int m = n / 3;map<char, int> mp;for (int i = 0; i < m; i++){for (int j = 0; j < m; j++){for (int p = 0; p < 3; p++){for (int q = 0; q < 3; q++){if (board[i*3+p][j*3+q] == '.') continue;if (mp.find(board[i*3+p][j*3+q]) == mp.end()){mp[board[i * 3 + p][j * 3 + q]] = 1;}else return false;}}mp.clear();}}return true;}};

isValidSudoku里边本来写的 return rowValid(board,n)&&colValid(board,n)&&blockValid(board,n),后来发现这样写可能拉高了平均复杂度，最后还是分开写了。

第二遍ac的代码（只遍历一遍数组，用三个二维数组分别存放行、列、九宫格中元素出现的情况）：

class Solution{public:bool isValidSudoku(vector<vector<char> > &board){vector<vector<bool> > rows(9, vector<bool>(9, false));vector<vector<bool> > cols(9, vector<bool>(9, false));vector<vector<bool> > blocks(9, vector<bool>(9, false));for (int i = 0; i < 9; i++){for (int j = 0; j < 9; j++){if (board[i][j] == '.') continue;int num = board[i][j] - '1';if (rows[i][num] || cols[j][num] || blocks[i - i % 3 + j / 3][num])return false;rows[i][num] = cols[j][num] = blocks[i - i % 3 + j / 3][num] = true;}}return true;}};

这段代码比较简洁一点，九宫格的坐标换算可能比较难想。前后两个方法的效率差距还挺大的。第一个120ms AC，第二个96ms AC。个人感觉是不是map的find方法浪费了时间？