Leetcode 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.."]
]

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
;
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。