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leetcode - N-Queens

时间:2015-09-20 22:02:49      阅读:169      评论:0      收藏:0      [点我收藏+]

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

技术分享

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.."]
]

class Solution {
private:
    std::vector<std::vector<std::string> > res;
public:
    std::vector<std::vector<std::string> > solveNQueens(int n) {
        std::vector<std::string>cur(n, std::string(n,'.'));
        dfs(cur, 0);
        return res;
    }
    void dfs(std::vector<std::string> &cur, int row)
    {
        if(row == cur.size())
        {
            res.push_back(cur);
            return;
        }
        for(int col = 0; col < cur.size(); col++)
            if(isValid(cur, row, col))
            {
                cur[row][col] = 'Q';
                dfs(cur, row+1);
                cur[row][col] = '.';
            }
    }

	//推断cur[row][col]位置放置皇后。是否合法。
    bool isValid(std::vector<std::string> &cur, int row, int col)
    {
        //列
        for(int i = 0; i < row; i++)
            if(cur[i][col] == 'Q')return false;
        //右对角线
        for(int i = row-1, j=col-1; i >= 0 && j >= 0; i--,j--)
            if(cur[i][j] == 'Q')return false;
        //左对角线
        for(int i = row-1, j=col+1; i >= 0 && j < cur.size(); i--,j++)
            if(cur[i][j] == 'Q')return false;
        return true;
    }
};


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leetcode - N-Queens

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原文地址:http://www.cnblogs.com/hrhguanli/p/4824374.html

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