一、问题描述
判断一个不完整的数独是否正确。
判断数独是否正确的条件：1、每行的数不重复。2、没列的数不重复。3、划分成的9个3*3的块中没有数重复。
例子看下面代码中的测试用例。
二、问题解决
思路一：分3次循环来解决，第一次横向遍历，第二次纵向遍历，第三次以块遍历。每一次遍历一行、一列、或是一块的时候使用map记录每个数出现的次数，当出现两次的时候返回false
bool isValidSudoku(vector<vector<char>>& board) { for (auto i : board) { map<char, int> map = { { ‘1‘,0 },{ ‘2‘,0 },{ ‘3‘,0 },{ ‘4‘,0 }, { ‘5‘,0 },{ ‘6‘,0 },{ ‘7‘,0 },{ ‘8‘,0 },{ ‘9‘,0 }, }; for (auto j : i) { if (j == ‘.‘) continue; if (map[j] == 0) map[j]++; else return false; } map = { { ‘1‘,0 },{ ‘2‘,0 },{ ‘3‘,0 },{ ‘4‘,0 }, { ‘5‘,0 },{ ‘6‘,0 },{ ‘7‘,0 },{ ‘8‘,0 },{ ‘9‘,0 }, }; } for (int i = 0; i < 9; i++) { map<char, int> map = { { ‘1‘,0 },{ ‘2‘,0 },{ ‘3‘,0 },{ ‘4‘,0 }, { ‘5‘,0 },{ ‘6‘,0 },{ ‘7‘,0 },{ ‘8‘,0 },{ ‘9‘,0 }, }; for (int j = 0; j < 9; j++) { if (board[j][i] == ‘.‘) continue; if (map[board[j][i]] == 0) map[board[j][i]]++; else return false; } map = { { ‘1‘,0 },{ ‘2‘,0 },{ ‘3‘,0 },{ ‘4‘,0 }, { ‘5‘,0 },{ ‘6‘,0 },{ ‘7‘,0 },{ ‘8‘,0 },{ ‘9‘,0 }, }; } for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { map<char, int> map = { { ‘1‘,0 },{ ‘2‘,0 },{ ‘3‘,0 },{ ‘4‘,0 }, { ‘5‘,0 },{ ‘6‘,0 },{ ‘7‘,0 },{ ‘8‘,0 },{ ‘9‘,0 }, }; for (int m = 0; m < 3; m++) { for (int n = 0; n < 3; n++) { int posx = m + i * 3; int posy = n + j * 3; if (board[posx][posy] == ‘.‘) continue; if (map[board[posx][posy]] == 0) map[board[posx][posy]]++; else return false; } } map = { { ‘1‘,0 },{ ‘2‘,0 },{ ‘3‘,0 },{ ‘4‘,0 }, { ‘5‘,0 },{ ‘6‘,0 },{ ‘7‘,0 },{ ‘8‘,0 },{ ‘9‘,0 }, }; } } return true; } int main() { vector<vector<char>> v = {{‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘5‘, ‘.‘, ‘.‘, ‘1‘,‘.‘}, { ‘.‘, ‘4‘, ‘.‘, ‘3‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘ }, { ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘3‘, ‘.‘, ‘.‘, ‘1‘ }, { ‘8‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘2‘, ‘.‘ }, { ‘.‘, ‘.‘, ‘2‘, ‘.‘, ‘7‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘ }, { ‘.‘, ‘1‘, ‘5‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘ }, { ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘2‘, ‘.‘, ‘.‘, ‘.‘ }, { ‘.‘, ‘2‘, ‘.‘, ‘9‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘ }, { ‘.‘, ‘.‘, ‘4‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘ } }; cout << isValidSudoku(v) << endl; system("pause"); return 0; }
思路二：更简洁方便的解决
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; }
三、问题总结
在判断是否出现重复的时候使用9*9的数组来表示，而不是使用map。
遍历一遍就ok了。
在利用i和j计算块内索引的时候很巧妙，记住这种方法。
大神的代码就是厉害。
