[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.."]
]

用2个栈stack模拟递归bfs方法，发现stack比递归速度快很多，递归的方法在9皇后时速度特别慢，通不过，用stack就会Accept！

class Solution {
public:
vector<vector<string> > result;

vector<vector<string> > solveNQueens(int n) {
vector<int> board(n+1,0);
stack<vector<int> > s;
s.push(board);
return  totalN(n,s);

}
private:
vector<vector<string> > totalN(int n,stack<vector<int> > s){

int k = 1;
while(k<=n){
stack<vector<int> > s0;
while(!s.empty()){
vector<int> board = s.top();
vector<int>  board0(board);
s.pop();

for(int i=1;i<=n;i++){
board[k]=i;
if(isok(k,board)){
s0.push(board);
}
board = board0;
}

}
k++;
s = s0;
}

while(!s.empty()){
string s0(n,'.');
vector<string> vs(n,s0);
vector<int> board = s.top();
s.pop();
for(int i=1;i<=n;i++){
vs[i-1][board[i]-1]='Q';
}
result.push_back(vs);
}
return result;
}
bool isok(int k,vector<int> &board){//check whether the ktn Queen can be put down
for(int i=1;i<k;i++){
if(board[i]==board[k]||abs(i-k)==abs(board[i]-board[k]))
return false;
}
return true;
}
};