N-Queens

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

Given an integer n, return all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

For example,
There exist two distinct solutions to the 4-queens puzzle:

[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

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主要参考 http://blog.csdn.net/hackbuteer1/article/details/6657109

class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        int i;
        //初始化
        for (i=0; i<n*n; i++) {
            _queue_arr[i] = INIT_VAL;
        }
        vector<vector<string>> res;
        int row = 0;
        int col = 0;
        //遍历每一行
        while (row < n) {
            //遍历每一列
            while (col < n) {
                //如果冲突,就计算下一列
                if (is_conflict(row, col, n)) {
                    col++;
                } else {
                    _queue_arr[row] = col;
                    col = 0;
                    break;
                }
            }
            //遍历完每一列后,检查是否成功放置皇后
            //如果失败
            if (col == n) {
                //如果当前行是第0行了,就说明所有的jie都被求出了
                if (0 == row) {
                    break;
                } else {
                    //如果不是,就从上一行的下一列开始放置皇后,并进行判断
                    row--;
                    col = _queue_arr[row];
                    col++;
                    _queue_arr[row] = INIT_VAL;
                    continue;
                }
                //如果当前行是最后一行了,就输出
            } else if (row == n - 1) {
                handle_result(res, n);
                col = _queue_arr[row] + 1;
                _queue_arr[row] = INIT_VAL;
                continue;
                //不是最后一行,就计算下一行
            } else {
                row++;
            };
        }
        return res;
    }

private:
    /*
     * 判断是否冲突,主要检测方式是 ,如果冲突
     * queue_arr[i] == queue_arr[j]
     * |queue_arr[i] -  queue_arr[j]| = |i - j|
     *
     * @param row: 当前行
     * @param col: 当前列
     * @param n: 所有的行数和列数
     */
    bool is_conflict(int row, int col, int n) const{
        int i;
        for (i=0; i<n; i++) {
            if (_queue_arr[i] == col || abs(_queue_arr[i] - col) == abs(i - row)) {
                return true;
            }
        }
        return false;
    }

    /*
     * 处理结果
     * @param res_str 结果集合
     * @param n  行数
     */
    void handle_result(vector<vector<string>> &res_str, int n) {
        vector<string> slu_str;
        int i,j;
        for (i=0; i<n; i++) {
            string tmp = "";
            for (j=0; j<n; j++) {
                if (_queue_arr[i] == j) {
                    tmp += "Q";
                } else {
                    tmp += ".";
                }
            }
            slu_str.push_back(tmp);
        }
        res_str.push_back(slu_str);
    }



private:
    int _queue_arr[1000];
    const static int INIT_VAL = -10000;

};