52. N-Queens II[hard]

Follow up for N-Queens problem.

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

经典N皇后问题，直接dfs即可

class Solution {public:    int ans;        bool check(int x1, int y1 , int x2 , int y2)    {        if(x1 == x2 || y1 == y2)            return false;        if( (x1-x2) == (y1-y2) || (x1-x2) == (y2-y1))            return false;        return true;    }        void dfs(int now , vector<pair<int,int> > &m, int n)    {        if(now==n+1)        {            ans++;            return;        }        for(int i=1;i<=n;i++)        {            bool flag = true;            for(int j=0;j<m.size() && flag;j++)            {                int x = m[j].first , y = m[j].second;                if(!check(now,i,x,y))                    flag=false;            }            if(flag)            {                m.push_back(make_pair(now,i));                dfs( now + 1 , m , n );                m.pop_back();            }        }    }    int totalNQueens(int n)    {        ans = 0;        vector<pair<int,int> > m;        dfs(1,m,n);        return ans;    }};