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.

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

[

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

  "...Q",

  "Q...",

  "..Q."],



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

  "Q...",

  "...Q",

  ".Q.."]

]
思路：八皇后问题一直是一个经典的回溯算法，递归算法解决。这道题受字符串的全排列的启发——定义一个数组data[n],数组中第i个数组表示位于第i行的皇后的列号。先把数组data的n个数字初始化分别用0~n-1初始化，接下来就是对数组data做全排列。我们只需要判断每一个排列对应的n个皇后是不是在同一对角线上，也就是对于数组的两个下标i和j，是不是j-i=abs(data[j]-data[i]).

class Solution {

public:

    bool check(vector<int> &data,int n)

    {

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

        {

            for(int j=i+1;j<n;j++)

            {

                if((j-i)==abs(data[j]-data[i]))

                {

                    return false;

                }

            }

        }

        return true;

    }

    void Permutation(vector<vector<string> > &result,vector<int> &data,int pBegin,int n)

    {

        if(pBegin==n)

        {

            if(true==check(data,n))

            {

                vector<string> pTemp;

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

                {

                    string row(n,'.');

                    row[data[i]]='Q';

                    pTemp.push_back(row);

                }

                result.push_back(pTemp);

            }

        }

        else

        {

            for(int i=pBegin;i<n;i++)

            {

                swap(data[pBegin],data[i]);

                Permutation(result,data,pBegin+1,n);

                swap(data[pBegin],data[i]);

            }

        }

    }

    void Permutation(vector<vector<string> > &result,int n)

    {

        vector<int> data(n);//

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

            data[i]=i;

        Permutation(result,data,0,n);

    }

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

        vector<vector<string> > result;

        result.clear();

        Permutation(result,n);

        return result;

    }

};