051.N-Queens

Posted 2020-08-07

经典的八皇后问题

 1 class Solution {
 2 public:
 3     vector<vector<string>> solveNQueens(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;
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 n
21                 )
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                 }
38                 else {
39                     checkPos(result, pos, isColumnSafe, isDiag1Safe, isDiag2Safe, i + 1, n);
40                 }
41                 isColumnSafe[j] = 1;
42                 isDiag1Safe[i - j + n - 1] = 1;
43                 isDiag2Safe[i + j] = 1;
44             }
45         }
46     }
47 };