运用回溯的思路，递归的编写解。
建立4个数组分别代表行、列、左上至右下对角线、左下至右上对角线（对于N*N的棋盘，通过数学归纳法易证，其有(2N-1)条对角线）。
对于左上至右下对角线横纵坐标的差一定，对于左下至右上对角线横纵坐标的和一定，由此即可判断某坐标在那一条对角线上。（注意：对于左上至右下对角线，横纵坐标差的取值在[-(N-1),N-1],需要加上（N-1）使序号非负）
每放置一个皇后便更改4个标记数组的值，使以后的皇后不能出现在同一行、列、对角线上。
时间复杂度：O（N^2）
C++代码：

class Solution {
	vector<vector<string>> result;
	vector<bool> diagTopLeftToBottomRight;
	vector<bool> diagTopRightToBottomLeft;
	vector<bool> col;
	vector<bool> row;
	vector<int>queen;
public:
	vector<vector<string>> solveNQueens(int n) {
		if (n == 1)
			return { {"Q"} };
		if (n < 4)
			return {};
		initial(n);
		nQueen(n, 0, queen);
		return result;
	}
	void initial(int n)
	{
		diagTopLeftToBottomRight.resize(2 * n - 1);
		diagTopRightToBottomLeft.resize(2 * n - 1);
		row.resize(n);
		col.resize(n);
		for (int i = 0; i < 2 * n - 1; i++)
		{
			diagTopLeftToBottomRight[i] = diagTopRightToBottomLeft[i] = true;
		}
		for (int i = 0; i < n; i++)
		{
			row[i] = col[i] = true;
		}

	}
	void pushAnswer(vector<int> queenPos) {
		vector<string> answer;
		answer.resize(queenPos.size());
		for (int i = 0;i<queenPos.size();i++)
		{
			for (int j = 0; j < queenPos.size(); j++)
			{
				answer[i].push_back((queenPos[i] == j) ? 'Q' : '.');
			}
		}
		result.push_back(answer);
	}
	bool judge(int x,int y ,int n)
	{
		if (row[x] == true && col[y] == true &&
			diagTopLeftToBottomRight[x - y + n - 1] == true &&
			diagTopRightToBottomLeft[x + y] == true)
		{
			return true;
		}
		return false;
	}
	void nQueen(int n, int now,vector<int> &queenPos)
	{
		if (now == n)
		{
			pushAnswer(queenPos);
			return;
		}
		for (int i = 0; i < n; i++)
		{
			if (judge(now, i,n))
			{
				row[now] = false;
				col[i] = false;
				diagTopLeftToBottomRight[now - i + n - 1] = false;
				diagTopRightToBottomLeft[now + i] = false;
				queenPos.push_back(i);
				nQueen(n, now + 1, queenPos);
				queenPos.pop_back();
				row[now] = true;
				col[i] = true;
				diagTopLeftToBottomRight[now - i + n - 1] = true;
				diagTopRightToBottomLeft[now + i] = true;
			}
		}
	}
};