LeetCode Everyday--52

Follow up for N-Queens problem.

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

[code]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。