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The idea is to use backtracking. In fact, the code below uses DFS, which involves backtracking in a recursive manner.
The idea is also very simple. Starting from the first row, try each column. If it does not induce any attack, move on to the next row based on the configurations of the previous rows. Otherwise, backtrack to the current row and try another selection of the column position. Once we reach the last row, add the current setting to a vector<vector<string> >.
The code below is referenced to this link, which records the positions of the queens using a nice 1d array like a[row] = col to indicate there is a queen at (row, col).
The code is as follows.
1 class Solution { 2 public: 3 vector<vector<string>> solveNQueens(int n) { 4 vector<vector<string> > queens; 5 vector<int> colPos(n, 0); 6 solve(colPos, n, 0, queens); 7 return queens; 8 } 9 private: 10 bool noAttack(vector<int>& colPos, int row, int col) { 11 for (int r = row - 1, ld = col - 1, rd = col + 1; r >= 0; r--, ld--, rd++) 12 if (colPos[r] == col || colPos[r] == ld || colPos[r] == rd) 13 return false; 14 return true; 15 } 16 vector<string> queenStr(vector<int>& colPos) { 17 int n = colPos.size(); 18 vector<string> queen(n, string(n, ‘.‘)); 19 for (int i = 0; i < n; i++) 20 queen[i][colPos[i]] = ‘Q‘; 21 return queen; 22 } 23 void solve(vector<int>& colPos, int n, int row, vector<vector<string> >& queens) { 24 if (row == n) { 25 queens.push_back(queenStr(colPos)); 26 return; 27 } 28 for (int col = 0; col < n; col++) { 29 colPos[row] = col; 30 if (noAttack(colPos, row, col)) 31 solve(colPos, n, row + 1, queens); 32 } 33 } 34 };
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4614567.html
