N-Queens

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

Given an integer n, return all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

Example:

Input: 4
Output: [
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],

["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.

class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        vector<vector<string>> solutions;
        vector<int> queens(n,-1);
        unordered_set<int> columns;//列
        unordered_set<int> diagonals1;//对角线1
        unordered_set<int> diagonals2;//对角线2
        backtrack(solutions,queens,n,0,columns,diagonals1,diagonals2);
        return solutions;
    }

    void backtrack(vector<vector<string>> &solutions,vector<int> &queens,int n,int row,unordered_set<int> &columns,unordered_set<int> diagonals1,unordered_set<int> diagonals2)
    {
        if(row == n){
            vector<string> board = generateBoard(queens,n);
            solutions.push_back(board);
        } else{
            for(int i=0;i<n;++i){
                if(columns.find(i)!= columns.end()){
                    continue;
                }
                int diagonal1 = row -i;
                if(diagonals1.find(diagonal1) != diagonals1.end()){
                    continue;
                }
                int diagonal2 = row +i;
                if(diagonals2.find(diagonal2) != diagonals2.end()){
                    continue;
                }
                queens[row] = i;
                columns.insert(i);
                diagonals1.insert(diagonal1);
                diagonals2.insert(diagonal2);
                backtrack(solutions,queens,n,row+1,columns,diagonals1,diagonals2);
                queens[row] = -1;
                columns.erase(i);
                diagonals1.erase(diagonal1);
                diagonals2.erase(diagonal2);
            }
        }
    }
    vector<string> generateBoard(vector<int> &queens,int n){
        vector<string> board;
        for(int i=0;i<n;++i)
        {
            string row = string(n,'.');
            row[queens[i]] = 'Q';
            board.push_back(row);
        }
        return board;
    }
};

注意：回溯算法，深度优先查找。