[LeetCode] 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.."]
]

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思路：递归，dfs，回溯法

 

首先实现了一个n皇后的dfs，X[i] 表示第i行所处的位置

#define N 4

vector<int> x;

bool checkTwoPoints(int i, int j, int xi, int xj) 
{
    //cout << "check i	" << i << endl;
    //cout << "check j	" << j << endl;
    if(xi == xj) // same column
        return false;
    if( abs(xi-xj) == abs(i-j)) // diag
        return false;
    return true;
}

bool check(vector<int> x, int n) // check x[n] and x[0 .. n-1]
{
    for(int i = 0; i < n; i++)
    {   
        if(!checkTwoPoints(i, n, x[i], x[n]))
            return false;
    }   
    return true;
}

void dfs(int n)
{
    if(n == N)
        printVector(x);

    for(int i = 0; i < N; i++)
    {
        x[n] = i;

        // check if x[n] is available
        if(check(x, n))
            dfs(n+1);
    }

}

int main()
{
    x.resize(N);
    dfs(0);
#if 0
    Solution sl;
    printVector (sl.plusOne(a));
    cout <<endl<< "==============" <<endl;
    a.clear();
#endif
    return 0;
}

然后考虑按照其格式打印字符串

class Solution {
    vector<int> x;
    vector< vector<string>  > m_res;

    bool checkTwoPoints(int i, int j, int xi, int xj)
    {
        //cout << "check i	" << i << endl;
        //cout << "check j	" << j << endl;
        if(xi == xj) // same column
            return false;
        if( abs(xi-xj) == abs(i-j)) // diag
            return false;
        return true;
    }

    bool check(vector<int> x, int n) // check x[n] and x[0 .. n-1]
    {
        for(int i = 0; i < n; i++)
        {
            if(!checkTwoPoints(i, n, x[i], x[n]))
                return false;
        }
        return true;
    }

    void dfs(int n)
    {
        if(n == x.size() )
        {
#if 1
            //printVector(x);

            string tmpStr;
            vector<string> strs;
            strs.resize(x.size());
            for(int i = 0; i < x.size(); i++)
                tmpStr += '.';

            for(int i = 0; i < x.size(); i++)
            {
                strs[i] = tmpStr;
                strs[i][x[i]] = 'Q';
            }
            m_res.push_back(strs);
            return;
#endif
        }

        for(int i = 0; i < x.size(); i++)
        {
            x[n] = i;

            // check if x[n] is available
            if(check(x, n))
                dfs(n+1);
        }

    }
    public:
    vector<vector<string> > solveNQueens(int n)
    {
        x.resize(n);
        dfs(0);
        return m_res;
    }
};