// 没有太大意义，和51题重复，只需要计数或者输出vector的大小即可
 1 class Solution { 2 public: 3     int totalNQueens(int n) { 4         vector<vector<string>> result; 5         vector<int> pos(n, -1); 6         vector<int> isColumnSafe(n, 1); 7         vector<int> isDiag1Safe(2 * n - 1, 1); // 主对角线 8         vector<int> isDiag2Safe(2 * n - 1, 1); // 斜对角线 9         checkPos(result, pos, isColumnSafe, isDiag1Safe, isDiag2Safe, 0, n);10         return result.size();11     }12 private:13 void checkPos(14     vector<vector<string>>& result,15     vector<int>& pos,16     vector<int>& isColumnSafe,17     vector<int>& isDiag1Safe,18     vector<int>& isDiag2Safe,19     int i,20     int n21 )22 {23     for (int j = 0; j < n; ++j) {24         bool flag = isColumnSafe[j] && isDiag1Safe[i - j + n - 1] && isDiag2Safe[i + j];25         if (flag) {26             isColumnSafe[j] = 0;27             isDiag1Safe[i - j + n - 1] = 0;28             isDiag2Safe[i + j] = 0;29             pos[i] = j;30             if (i == n - 1) {31                 vector<string> vec(n, string(n, ‘.‘));32                 int cnt = 0;33                 for (int k = 0; k < n; ++k) {34                     vec[cnt ++][pos[k]] = ‘Q‘;35                 }36                 result.push_back(vec);37                 for (int i = 0; i < n; ++i) {38                     cout << vec[i] << endl;39                 }40                 cout << endl;41             }42             else {43                 checkPos(result, pos, isColumnSafe, isDiag1Safe, isDiag2Safe, i + 1, n);44             }45             isColumnSafe[j] = 1;46             isDiag1Safe[i - j + n - 1] = 1;47             isDiag2Safe[i + j] = 1;48         }49     }50 }51 };