leetcode之n皇后问题

leetcode上有两个关于n皇后的问题，两个题目基本是一样的，只是第二个是把所有的排法求出来。n皇后最简单的就是用递归，每次判断一行的一个位置，如果合法，就判断下一行，不合法再判断下一个位置

N-Queens II

 

Follow up for N-Queens problem.

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

class Solution {
public:
    //由于每次都是遍历下一行，所以两个皇后的行肯定不同；由于hasUse用于判断当前列是否已经占用，所以这里之需要判断对角线的位置
    bool check(int row,int col,int n,vector queue)
    {
    	int i;
    	for(i = 0; i < row;i++)
    	{
    		if(abs(i - row) == abs(queue[i]-col))return false;
    	}
    	return true;
    }
    //n表示皇后的个数
    void totalNQueens(int row,int n,vector& queue,vector& hasUse,int& count)
    {
    	if(row == n)
    	{
    		count ++;
    		return;
    	}
    	int col;
    	for(col = 0;col < n;++col)
    	{
    		if(!hasUse[col] && check(row,col,n,queue))
    		{
			hasUse[col] = true;//表示第col列已经被使用
			queue[row] = col;//表示第row行选择的是第col列作为皇后
			totalNQueens(row+1,n,queue,hasUse,count);
			hasUse[col] = false;
			queue[row] = -1;
    		}
    	}
    }
    int totalNQueens(int n)
    {
    	int count = 0;
    	vector queue(n,-1);
    	vector hasUse(n,false);
    	totalNQueens(0,n,queue,hasUse,count);
    	return count;
    }
};

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.."]
]

class Solution {
public:
    bool check(int row,int col,int n,vector queue)
    {
    	int i;
    	for(i = 0; i < row;i++)
    	{
    		if(abs(i - row) == abs(queue[i]-col))return false;
    	}
    	return true;
    }
    
    void solveNQueens(int row,int n,vector& queue,vector& hasUse)
    {
    	if(row == n)//只有此处和上面不同，这里是把每一个位置保存到结果中而不是记录个数
    	{
    		vector solve;
    		int i,j;
    		for(i = 0;i < row;++i)
    		{
    			string s;
    			for(j = 0;j < queue[i];++j)s += '.';
    			s += 'Q';
    			for(++j;j < n;++j) s+= '.';
    			solve.push_back(s);
    		}
    		res.push_back(solve);
    		return;
    	}
    	int col;
    	for(col = 0;col < n;++col)
    	{
    		if(!hasUse[col] && check(row,col,n,queue))
    		{
    			hasUse[col] = true;
    			queue[row] = col;
    			solveNQueens(row+1,n,queue,hasUse);
    			hasUse[col] = false;
    			queue[row] = -1;
    		}
    	}
    }
    vector > solveNQueens(int n)
    {
    	vector queue(n,-1);
    	vector hasUse(n,false);
    	solveNQueens(0,n,queue,hasUse);
    	return res;
    }
private:
    vector > res;
};