[LeetCode] N-Queens N皇后问题

 

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.

For example,
There exist two distinct solutions to the 4-queens puzzle:

[

 [".Q..",  // Solution 1

  "...Q",

  "Q...",

  "..Q."],



 ["..Q.",  // Solution 2

  "Q...",

  "...Q",

  ".Q.."]

]

 

经典的N皇后问题，基本所有的算法书中都会包含的问题，经典解法为回溯递归，一层一层的向下扫描，需要用到一个pos数组，其中pos[i]表示第i行皇后的位置，初始化为-1，然后从第0开始递归，每一行都一次遍历各列，判断如果在该位置放置皇后会不会有冲突，以此类推，当到最后一行的皇后放好后，一种解法就生成了，将其存入结果res中，然后再还会继续完成搜索所有的情况，代码如下：

 

class Solution {

public:

    vector<vector<string> > solveNQueens(int n) {

        vector<vector<string> > res;

        vector<int> pos(n, -1);

        solveNQueensDFS(pos, 0, res);

        return res;

    }

    void solveNQueensDFS(vector<int> &pos, int row, vector<vector<string> > &res) {

        int n = pos.size();

        if (row == n) {

            vector<string> out(n, string(n, '.'));

            for (int i = 0; i < n; ++i) {

                out[i][pos[i]] = 'Q';

            }

            res.push_back(out);

        } else {

            for (int col = 0; col < n; ++col) {

                if (isValid(pos, row ,col)) {

                    pos[row] = col;

                    solveNQueensDFS(pos, row + 1, res);

                    pos[row] = -1;

                }

            }

        }

    }

    bool isValid(vector<int> &pos, int row, int col) {

        for (int i = 0; i < row; ++i) {

            if (col == pos[i] || abs(row - i) == abs(col - pos[i])) {

                return false;

            }

        }

        return true;

    }

};

 

此题还有非递归的解法，请参见网友JustDoIt的博客。