Leetcode N-Queens

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

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

Each solution contains a distinct board configuration of then-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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虽然不懂国际象棋，但它应该是可以攻击与它在同一行，同一列或对角线上的棋子。解题的方法就是递归大法。

代码如下：

class Solution {public:    vector<vector<string>> solveNQueens(int n) {        vector<vector<string>> result;        vector<string> answer;        int* visit = new int[n];        memset(visit,-1,sizeof(int)*n);        for(int i=0;i<n;i++)        {            string temp = string(n,'.');            temp[i] = 'Q';            visit[i] = 0;            answer.push_back(temp);            Queens(result,answer,visit,i,1,n);            visit[i] = -1;            answer.pop_back();        }        return result;    }        void Queens(vector<vector<string>>& result,vector<string>& answer,int* visit,int index,int count,int n)    {        if(count == n)        {            result.push_back(answer);            return;        }                bool flag;        for(int i =0;i<n;i++)        {            flag = true;            string temp = string(n,'.');            if(visit[i] != -1)                continue;            for(int j=0;j<n;j++)            {                if(visit[j] != -1)                {                    if(abs(count-visit[j]) == abs(i-j))                    {                        flag = false;                        break;                    }                }            }            if(flag == false)                continue;            temp[i] = 'Q';            visit[i] = count;            answer.push_back(temp);            Queens(result,answer,visit,i,count+1,n);            visit[i] = -1;            answer.pop_back();        }    }};

看着自己的代码闻到一股shit的味道，要使代码更加优雅整洁，修改后的代码如下：

class Solution {public:    vector<vector<string>> solveNQueens(int n) {        vector<vector<string>> result;        vector<string> answer(n,string(n,'.'));        Queens(result,answer,0,n);        return result;    }        bool valid(vector<string>& answer,int row,int col,int n)    {        for(int i=0;i<n;i++)            if(answer[row][i] == 'Q')                return false;                        for(int i=0;i<n;i++)            if(answer[i][col] == 'Q')                return false;                        for(int i=row,j=col;i>=0&&j>=0;i--,j--)            if(answer[i][j] == 'Q')                return false;                for(int i=row,j=col;i>=0&&j<n;i--,j++)            if(answer[i][j] == 'Q')                return false;                return true;            }        void Queens(vector<vector<string>>& result,vector<string>& answer,int row,int n)    {        if(row == n)        {            result.push_back(answer);            return;        }                for(int col =0;col<n;col++)        {            if(valid(answer,row,col,n))            {                answer[row][col] = 'Q';                Queens(result,answer,row+1,n);                answer[row][col] = '.';            }        }    }};

需要特别注意的是，在进行在斜线判断的时候，因为状态是从上一行往下一行传递，所以我们只需要判断45度和135度即可。