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.
Example:
Input: 4
Output: [
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],
["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.
Thoughts
- Classic N-Queens problem
- 使用经典的
backtracking方法来解决 - 每次放一个Queen，然后检查结果的合理性，不合理就退后一步，合理就继续进行
- 直到把所有的皇后都放置完毕
- 注意：
- 判断对角线是否有皇后时，要注意有两条对角线
- 因为每一行只能放置一个皇后，所以可以利用这个条件，放置一个皇后后就不用循环改行所有的位置了，因为必定是不正确的，从而降低时间复杂度
Code(JAVA)
List<List<String>> res;
public List<List<String>> solveNQueens(int n) {
char[][]board = new char[n][n];
res = new ArrayList();
//init board
for(int i = 0; i < n; i ++){
for(int j = 0; j < n; j++){
board[i][j] = '.';
}
}
placeQueens(board,0);
return res;
}
private void placeQueens(char[][]board,int curRow) {
int n = board.length;
if(curRow == n){
res.add(convertResult(board,n));
return;
}
for(int i = 0; i < n; i ++){
if(validQueen(board,curRow,i)) {
board[curRow][i] = 'Q';
placeQueens(board,curRow+1);
board[curRow][i] = '.';
}
}
}
private boolean validQueen(char[][]board,int row, int col){
for (int i = 0; i < row; ++i) {
if (board[i][col] == 'Q') return false;
}
for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; --i, --j) {
if (board[i][j] == 'Q') return false;
}
for (int i = row - 1, j = col + 1; i >= 0 && j < board.length; --i, ++j) {
if (board[i][j] == 'Q') return false;
}
return true;
}
private List<String> convertResult(char[][]board, int n) {
List<String> ans = new ArrayList();
for(int i = 0; i < n; i++) {
ans.add(String.valueOf(board[i]));
}
return ans;
}
