LeetCode Everyday--52

Follow up for N-Queens problem.

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

class Solution {public:    int totalNQueens(int n) {        this->count = 0;        this->column = vector<int>(n,0);        this->main_diagonal = vector<int>(2 * n - 1,0);        this->anti_diagonal = vector<int>(2 * n - 1,0);        vector<int> queens = vector<int>(n,0);        placeQueen(queens,0);//放置皇后        return count;    }private:    int count ;//解的个数    vector<int> column;//存放所占据的列    vector<int> main_diagonal;//存放所占据主对角线    vector<int> anti_diagonal;//存放所占据主对角线    //输出皇后位置    /*    void printQueens(vector<int> queens){        int n = queens.size();        for(int i =0;i < n;i++){            for(int j =0;j < n;j++){                if(queens[i] == j){                    cout<<"# ";                }else{                    cout<<"* ";                }            }            cout<<endl;        }        cout<<endl;    }*/    //放置皇后    void placeQueen(vector<int> &queens,int row){        int n = queens.size();         //找到一个解        if(row == n){            count++;            //printQueens(queens);//输出解            return ;        }        for(int i = 0;i < n;i++){            //所在列及对角线未被占据              if(column[i] == 0 && main_diagonal[row + i] == 0 && anti_diagonal[i - row + n - 1] == 0){                //queens[row] = i;//记录第row行皇后所在位置                //将皇后所在列及对角线值1                column[i] = 1;                main_diagonal[row + i] = 1;                anti_diagonal[i - row + n - 1] = 1;                placeQueen(queens,row + 1);                //queens[row] = 0;                column[i] = 0;                main_diagonal[row + i] = 0;                anti_diagonal[i - row + n - 1] = 0;            }        }    }};

跟之前的解法略有不同，但是递归的思想基本相同。之前用一个二维数组储存棋盘，现在只需要存储所占据的列，主副对角线，每当摆放一个皇后的时候检查对应的列，主对角线是否被占据，若没有则摆放皇后并向下递归。