51.多皇后问题

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问题描述：

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.."]]

测试代码（c++）：

class Solution {    void add_queen(vector<vector<string>>& result,vector<string> res,vector<vector<bool> > index,int n,int row)    {        string s(n,'.');        if(row==n)        {            result.push_back(res);            return;        }        for(int i=0;i<n;i++)        {            if(index[row][i])                continue;            s[i] = 'Q';            res.push_back(s);            add_queen(result,res,edit(index,row+1,i),n,row+1);            s[i] = '.';            res.pop_back();        }    }    vector<vector<bool> > edit(vector<vector<bool> > index,int row,int i)    {        for(int j=row;j<index.size();j++)        {            index[j][i] = true;            if(i-j+row-1>=0)            {                index[j][i-j+row-1] = true;            }            if(i+j-row+1<index.size())            {                index[j][i+j-row+1] = true;            }        }        return index;    }public:    vector<vector<string>> solveNQueens(int n) {        vector<vector<string>> result;        vector<string> res;        vector<vector<bool> > index(n,vector<bool>(n,false));        add_queen(result,res,index,n,0);         return result;    }};

性能：

参考答案（c++）：

class Solution {public:    int abs(int x, int y)    {        return x>y ? x-y:y-x;    }    bool isSol(vector<int> & nums, int next)    {        for(int row=0; row<next; row++)        {            if(nums[row] == nums[next])                return false;            if(abs(next,row) == abs(nums[next],nums[row]))                return false;        }        return true;    }    void npossibility(vector<int> & pos, int index, vector<vector<int>> & result){    for (int i = 0; i<pos.size(); i++)    {        pos[index] = i;        if (isSol(pos, index))        {            if (index == pos.size() - 1)                result.emplace_back(pos);            else                 npossibility(pos, index + 1, result);        }    }}    vector<vector<string>> solveNQueens(int n) {        vector<vector<int>> result ;         vector<int> pos (n, 0);        npossibility(pos, 0 , result);        vector<vector<string>> strresult ;        string dotstr (n,'.');        vector<string> str (n, dotstr);        for(auto r:result)        {            vector<string> temp (str);            for(int row=0; row<r.size(); row++)            {                temp[row][r[row]] = 'Q';            }            strresult.emplace_back(temp);        }        return strresult;    }};

性能：

参考答案（python）：

    def DFS(queens, xy_dif, xy_sum):        p = len(queens)        if p==n:            result.append(queens)            return None        for q in range(n):            if q not in queens and p-q not in xy_dif and p+q not in xy_sum:                 DFS(queens+[q], xy_dif+[p-q], xy_sum+[p+q])      result = []    DFS([],[],[])    return [ ["."*i + "Q" + "."*(n-i-1) for i in sol] for sol in result]

性能：