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.."] ]
经典皇后问题,递归回溯搜索。
vector<vector<string> > res;
bool CanPut(vector<string>& tmp, int row, int col, int n)
{
int i=0, j=0;
//row行,col列有'Q'返回false
for(i=0; i<row; i++)
{
if(tmp[i][col] == 'Q')
return false;
}
//row行,col列的左斜对角有'Q'返回false
for(i=row-1, j=col-1; i>=0 && j>=0; i--, j--)
{
if(tmp[i][j] == 'Q')
return false;
}
//row行,col列的右斜对角有'Q'返回false
for(i=row-1, j=col+1; i>=0 && j<n; i--, j++)
{
if(tmp[i][j] == 'Q')
return false;
}
return true;
}
void solve(vector<string>& tmp, int row, int n)
{
if(row==n)
{
res.push_back(tmp);
return ;
}
for(int col=0; col<n; col++)
{
string s(n, '.');
if(CanPut(tmp, row, col, n))
{
//如果在row行,col列可以放‘S’
s[col] = 'Q';
tmp.push_back(s);
solve(tmp, row+1, n);
//回溯,递归中很重要的一步
//如果不满足,则回退一步搜索
tmp.pop_back();
}
}
}
vector<vector<string> > solveNQueens(int n)
{
if(n == 0)
return res;
vector<string> tmp;
solve(tmp, 0, n);
return res;
}
N皇后问题求解
本文介绍了一个经典的递归回溯算法解决N皇后问题的方法。通过递归地放置皇后并检查是否冲突来寻找所有可能的解决方案。展示了如何使用C++实现这一算法,并给出了一个4皇后问题的例子。
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