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 , 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
     
      
       > solveNQueens(int n) { 4         vector
       
        
          > queens; 5         vector
         
           colPos(n, 0); 6         solve(colPos, n, 0, queens); 7         return queens; 8     } 9 private:10     bool noAttack(vector
          
           & 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
           
             queenStr(vector
            
             & colPos) {17 int n = colPos.size();18 vector
             
               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
              
               & colPos, int n, int row, vector
               
                
                  >& 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 };
                
               
              
             
            
           
          
         
        
       
      
     

Well, if you have solved N-Queens II, you may know that problem has a nice bit-manipulation solution (you may refer to ). In fact, that bit-manipulation idea can also be used to solve this problem. The code is as follows.

1 class Solution { 2 public: 3     vector
     
      
       > solveNQueens(int n) { 4         vector
       
        
          > queens; 5         vector
         
           queen; 6         int limit = (1 << n) - 1; 7         solve(0, 0, 0, n, limit, queen, queens); 8         return queens; 9     }10 private:11     void solve(int hProj, int lProj, int rProj, int n, int limit, vector
          
           & queen, vector
           
            
              >& queens) {12 if (hProj == limit) {13 queens.push_back(queen);14 return;15 }16 int pos = ~(hProj | lProj | rProj);17 for (int i = 0; i < n; i++) {18 int p = (1 << i);19 if (pos & p) {20 string line(n ,'.');21 line[i] = 'Q';22 queen.push_back(line);23 solve(hProj | p, (lProj | p) << 1, (rProj | p) >> 1, n, limit, queen, queens);24 queen.pop_back();25 }26 }27 }28 };