51. N-Queens

Posted 2021-01-13 ruruozhenhao

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.

 AC code:

class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        int res = 0;
        vector<vector<string>> v;
        vector<string> nqueens(n, string(n, ‘.‘));
        solve(v, nqueens, n, 0);
        return v;
    }
    
    void solve(vector<vector<string>>& v, vector<string>& nqueens, int n, int row) {
        if (row == n) {
            v.push_back(nqueens);
            return;
        }
        for (int col = 0; col < n; ++col) {
            if (judge(nqueens, n, row, col)) {
                nqueens[row][col] = ‘Q‘;
                solve(v, nqueens, n, row+1);
                nqueens[row][col] = ‘.‘;
            }
        }
    }

    bool judge(vector<string> nqueens, int n, int x, int y) {
        // up and down
        for (int i = 0; i < n && i != x; ++i) {
            if (nqueens[i][y] == ‘Q‘) return false;
        }
        // right and left
        for (int i = 0; i < n && i != y; ++i) {
            if (nqueens[x][i] == ‘Q‘) return false;
        }
        // left up
        for (int i = x-1, j = y-1; i >= 0 && j >= 0; --i, --j) {
            if (nqueens[i][j] == ‘Q‘) return false;
        }
        // right up
        for (int i = x-1, j = y+1; i >= 0 && j < n; --i, ++j) {
            if (nqueens[i][j] == ‘Q‘) return false;
        }
        return true;
    }
};

Runtime: 32 ms, faster than 8.09% of C++ online submissions for N-Queens.