Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

 

思路：和 N-queens 一样，不过只统计个数而言，更加简单了

class Solution {    vector
     
       x;    int m_res;    bool checkTwoPoints(int i, int j, int xi, int xj)    {        //cout << "check i\t" << i << endl;        //cout << "check j\t" << j << endl;        if(xi == xj) // same column            return false;        if( abs(xi-xj) == abs(i-j)) // diag            return false;        return true;    }    bool check(vector
      
        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() )        {               //printVector(x);                m_res ++;             return;        }           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:    int totalNQueens(int n)    {        x.resize(n);        m_res = 0;        dfs(0);        return m_res;    }};