36. Valid Sudoku, leetcode

题目：

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.

思路及代码：

class Solution {public:    bool isValidSudoku(vector<vector<char>>& board) {bool ans1, ans2, ans3;for (auto row : board){vector<char> nine1;for (char a : row)if (a != '.') nine1.push_back(a);if (!isValid(nine1))return false;}for (int i = 0; i < 9; i++){vector<char> nine2;for (int j = 0; j < 9; j++)if (board[j][i] != '.') nine2.push_back(board[j][i]);//易错点 是（j， i）而不是（i，j）if (!isValid(nine2))return false;}for (int i = 0; i < 9; i++){vector<char> nine3;int x = (i / 3) * 3, y = (i % 3 ) * 3;//易错点 不应该 -1 了for (int p = x; p < x + 3; p++)//易错点 应该是x+3 而不是 3for (int q = y; q < y + 3; q++)if (board[p][q] != '.') nine3.push_back(board[p][q]);if (!isValid(nine3))return false;}return true;}    bool isValid(vector<char> nine)    {                unordered_map<char,int> pool;        for(char a : nine)        {            if(pool.count(a))            return false;            else             pool[a]++;        }        return true;    }};

更简便的做法：上面的用到了hash map，下面的直接将下标当做hash table里的key，用数组代替了

Three flags are used to check whether a number appear.

used1: check each row

used2: check each column

used3: check each sub-boxes

class Solution{public:    bool isValidSudoku(vector<vector<char> > &board)    {        int used1[9][9] = {0}, used2[9][9] = {0}, used3[9][9] = {0};                for(int i = 0; i < board.size(); ++ i)            for(int j = 0; j < board[i].size(); ++ j)                if(board[i][j] != '.')                {                    int num = board[i][j] - '0' - 1, k = i / 3 * 3 + j / 3;                    if(used1[i][num] || used2[j][num] || used3[k][num])                        return false;                    used1[i][num] = used2[j][num] = used3[k][num] = 1;                }                return true;    }};