51. N-Queens

题目

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.."]
]

解析

• 经典dfs+回溯方法

// add 51. N-Queens
class Solution_51 {
public:
	bool isValid(vector<string> &vec,int i,int j)
	{
		//判断当前放置位置（i，j）是否合理
		for (int row = 0; row < i;row++) //判断同一列列
		{
			if (vec[row][j]=='Q') // vec[i][j]
			{
				return false;
			}
		}
		//判断对角线45°
		for (int row = i-1, col = j-1; row >= 0 && col>=0;row--,col--)
		{
			if (vec[row][col] == 'Q')
			{
				return false;
			}
		}
		//判断对角线135%
		for (int row = i-1, col = j + 1; row >= 0 && col < vec.size();row--,col++) //写法： row >= 0, col < vec.size() 错误
		{
			if (vec[row][col] == 'Q')
			{
				return false;
			}
		}
		return true;
	}

	void solveNQueensHelp(vector<vector<string>> &vecs,vector<string> &vec,int row,int n)
	{
		if (row==n)
		{
			vecs.push_back(vec);
			return;
		}

		for (int col = 0; col < n;col++)
		{
			if (isValid(vec,row,col))
			{
				vec[row][col] = 'Q';
				solveNQueensHelp(vecs, vec, row + 1, n);
				vec[row][col] = '.';
			}
		}
		return;
	}

	vector<vector<string>> solveNQueens(int n) {

		vector<string> vec(n,string(n,'.')); 

		vector<vector<string> > vecs; //所有解

		solveNQueensHelp(vecs,vec,0,n);
		
		return vecs;
	}
};

题目来源

• 51. N-Queens
• N皇后问题--回溯法