题目:
解答:
只需要判断现在已有的元素是否有冲突,不需要判断是否可解。
分别遍历行 列 和9宫格 用map判断是否存在重复元素。
代码:
class Solution { public: bool isValidSudoku(vector<vector<char> > &board) { int n = 9; map<char, bool> have; for (int i = 0; i < n; i++) { have.clear(); for (int j = 0; j < n; j++) { if (board[i][j] != '.') { if (have.find(board[i][j]) == have.end()) have[board[i][j]] = true; else return false; } } } for (int i = 0; i < n; i++) { have.clear(); for (int j = 0; j < n; j++) { if (board[j][i] != '.') { if (have.find(board[j][i]) == have.end()) have[board[j][i]] = true; else return false; } } } for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { have.clear(); for (int k = 3 * i; k < 3 * i + 3; k++) { for (int m = 3 * j; m < 3 * j + 3; m++) { if (board[k][m] != '.') { if (have.find(board[k][m]) == have.end()) have[board[k][m]] = true; else return false; } } } } } return true; } };