N-Queens II

Follow up for N-Queens problem.

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

class Solution {public:void print(int queue[], int n, vector<vector<string>> &qipan) {        int i, j;        /*for(i = 0; i < n; i++) {            printf("row :%d, col :%d\n", i , queue[i]);        } */         vector<string> method;        for(i = 0; i < n; i++) {            string str(n,'.');            str[queue[i]] = 'Q';              method.push_back(str);        }            //printf("\n");        qipan.push_back(method);        //printf("\n");    }    void testQueue(int queue[], int row, int n, vector<vector<string>> &qipan) {        int j, k;        if(row >= n) print(queue, n, qipan);        else {            for(j = 0; j < n; j++) {                queue[row] = j;                for(k = 0; k < row; k++) {                    if((queue[k] - queue[row]) * (fabs(queue[k] - queue[row]) - fabs(k - row)) == 0) {                        break;                     }                    if(k == row - 1) testQueue(queue, row + 1, n, qipan);                }            }        }    }    int totalNQueens(int n) {        vector<vector<string>> qipan;        int queue[n];        int i;        for(i = 0; i < n; i++ ) queue[i] = -1;        for(i = 0; i < n; i++) {            //printf("method :\n");            queue[0] = i;            testQueue(queue, 1, n, qipan);        }        return qipan.size();    }  };