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.."] ]
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題目鏈接：https://leetcode.com/problems/n-queens/
題目大意：N皇后問題，求出所有滿足N皇后的解法。
思路：深度搜索，寬度搜索或者回溯。
參考代碼：
class Solution {
public:
vector<vector<string>> solveNQueens(int n) {
vector < vector <string> > ans ;
if ( n == 0 )
{
vector <string> temp ;
ans.push_back ( temp ) ;
return ans ;
}
for ( int i = 0 ; i < n ; i ++ )
{
vector < pair <int,int> > v ;
v.push_back ( make_pair ( 0 , i ) ) ;
queue < vector < pair <int,int> > > q ;
q.push ( v ) ;
while ( ! q.empty() )
{
v = q.front() ;
q.pop() ;
if ( v.size() == n )
{
ans.push_back ( buildBoard ( n , v ) ) ;
continue ;
}
int index = v.back().first + 1 ;
for ( int j = 0 ; j < n ; j ++ )
{
if ( isValid ( n , v , index , j ) )
{
vector < pair <int,int> > temp = v ;
temp.push_back ( make_pair ( index , j ) ) ;
if ( index == n - 1 )
ans.push_back ( buildBoard ( n , temp ) ) ;
else
q.push ( temp ) ;
}
}
}
}
return ans ;
}
private :
bool isValid ( int n , const vector < pair <int,int> >& v , int x , int y )
{
for ( int i = 0 ; i < v.size() ; i ++ )
{
if ( x == v[i].first || y == v[i].second || x - v[i].first == y - v[i].second || x - v[i].first == v[i].second - y )
return false ;
}
return true ;
}
vector <string> buildBoard ( int n , const vector < pair <int,int> >& v )
{
string s ( n , '.' ) ;
vector <string> b ( n , s ) ;
for ( int i = 0 ; i < v.size() ; i ++ )
{
b[v[i].first][v[i].second] = 'Q' ;
}
return b ;
}
};