51. N-Queens
The n-queens puzzle is the problem of placing n queens on an n x n chessboard such that no two queens attack each other.
Given an integer n, return all distinct solutions to the n-queens puzzle. You may return the answer in any order**.
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 1:
Input: n = 4
Output: [[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above
Example 2:
Solution:
class Solution {
public List<List<String>> solveNQueens(int n) {
List<List<String>> result = new ArrayList<List<String>>();
char[][] emptyBoard = new char[n][n];
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
emptyBoard[i][j] = '.';
}
}
Set<Integer> cols = new HashSet<Integer>();
Set<Integer> diagnoals = new HashSet<Integer>();
Set<Integer> antiDiagnoals = new HashSet<Integer>();
helper(0, cols, diagnoals, antiDiagnoals, emptyBoard, n, result);
return result;
}
private static void helper(int row, Set<Integer> cols, Set<Integer> diagnoals, Set<Integer> antiDiagnoals, char[][] board, int n, List<List<String>> result){
if (row == n){
result.add(transfer(board, n));
}
for (int col = 0; col < n; col++){
int currDiagnoal = row - col;
int currAntiDiagnoal = row + col;
if (cols.contains(col) || diagnoals.contains(currDiagnoal) || antiDiagnoals.contains(currAntiDiagnoal)){
continue;
}
cols.add(col);
diagnoals.add(currDiagnoal);
antiDiagnoals.add(currAntiDiagnoal);
board[row][col] = 'Q';
helper(row + 1, cols, diagnoals, antiDiagnoals, board, n, result);
cols.remove(col);
diagnoals.remove(currDiagnoal);
antiDiagnoals.remove(currAntiDiagnoal);
board[row][col] = '.';
}
}
private static List<String> transfer(char[][] board, int size){
List<String> subSolutions = new ArrayList<String>();
for (int i = 0; i < size; i++){
String row = new String(board[i]);
subSolutions.add(row);
}
return subSolutions;
}
}
// TC: O(n!)
// SC: O(n^2
)
N = 4, there are two ways of putting 4 queens:
[1, 3, 0, 2] --> the Queen on the first row is at y index 1, the Queen on the second row is at y index 3, the Queen on the third row is at y index 0 and the Queen on the fourth row is at y index 2.
[2, 0, 3, 1] --> the Queen on the first row is at y index 2, the Queen on the second row is at y index 0, the Queen on the third row is at y index 3 and the Queen on the fourth row is at y index 1.
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | |
|---|---|---|---|---|---|---|---|---|
| 0 | Q0 | | | / | |||||
| 1 | | | Q1 | / | |||||
| 2 | ` | | | / | Q2 | ||||
| 3 | ` | | | / | Q3 | ||||
| 4 | — | — | Q4 | — | — | — | — | — |
| 5 | Q5 | / | | | ` | ||||
| 6 | / | | | ` | Q6 | ||||
| 7 | | | Q7 | ` |
Key Insight： 每一行有且仅有一个queen
- How many levels in the recursion tree
8 levels, consider which col should this queen put
- How many states can have from each node?
at most(8 branches)
Q0 可以放在 0 1 2 3 4 5 6 7
同样的Q1 也可以方法在 0 1 2 3 4 5 6 7
...
Q7: 0 1 2 3 4 5 6 7
Time: 8^8 -> N^n -> N!
这是不考虑剪枝的情况下，会有8^8
Time = O(n^n * n)
Space = O(n)
-> optimized to (对列做减枝) O(n! * n)
int[A] stores the current solution on each row
index 0 1 2 3 4 5 6 7
A[i] 1 3 5 7 2 0 6 4
A[1]: 1 means the 1st queen is put on 1st column of the 1st row
A[3]: 3 means the 2nd queen is put on the 3rd column of the 2nd row
current_row: the current row we are interesting a new queen
Base case: The last row is done, 0 row left
recursive rule: If current position(i, j) is valid -> go to the next row: (i + 1)
伪代码
// index 0 1 2 3 4
// int A[N] stores the current solution on each row, A[8] = {1,3,5,7,4}
// 1 means the queen 0 is put on 1st column of the 1st row
// 3 means the queen 1 is put on the 3rd column of the 2nd row
// current_row, the current row we are inserting a new queen
void eightQueen(int[] A , cur, int current_row == 4){
//已经放的结果 //当前层数
if (current_row == N){
// base case
// print A[];
return;
}
for (int i = 0; i < 8; i++){
// we can try N columns to insert a new queen on this row
A[current_row] = i; // i is the column number
// check whether this configuration is valid or not
// using a helper function to check whether A[0. ... . current_row-1] conflicts the
// current queen inserted or not;
if (pass the check){
// invalide case: duplicate column index or slope == 1/-1
eightQueen(A, current_row + 1); // recursive rule
}
}
}
candidate: (4, 4) -> row = cur.size() col= i
如何check? -> check 这个和前面的每个点共存 -> check这个点和前面的每个点都不在同一行, 同一列, 同一对角线
同一行: 每行放一个, 不需要check
同一列:
同一对对角线:
(x1, y1), (x2, y2) -> check他们的斜率绝对值 == 1 -> |y2 - y1 | / |x2 - x1| == 1
private boolean valid(List<Integer> cur, int column){
int row = cur.size();
for (int i = 0; i < cur.size(); i++){
if (cur.get(i) == column || Math.abs(cur.get(i) - column) == row - i){
return false;
}
}
return true;
}
思考题 : If there are obstacles on the board
扫描图中的每一行, 遇到一个障碍就每每一行分成小的子行 -> 每个subrow能放1个queen
How to draw the recursion tree ??????
sub_row_0: [0] [0] - [0] [1] two cells
sub_row_1: [0][3] - [0][3]
sub_row_2: [0] [5] - [0] [7]
sub_row_3: [1] [0] - [1] [7]
....
sub_row_12 .....
Assume that we have 13 sub-rows:
hint: 每个subrow一定会放一个queen吗? 不一定
root: 0 queen has been put at the begining
/ | \
L0 sub_row_0: Q0 not put here Q0 put [0] [0] Q0 put[0] [1]
/ \ / \
L1 Q0 not put here Q0[0] [3] Q1 not put here Q1[0] [3]
L2
L3
...
L12
in more details, your signature of the function should be like
void DFS_EightQueen_WithObstacle(int[][] input, int indexQueen (当前处理哪个queen), int indexRow, (当前处理哪个子行) ....)
class Solution {
public List<List<String>> solveNQueens(int n) {
int size = n;
List<List<String>> solutions = new ArrayList<List<String>>();
// [[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
// 0 1 2 3
//0 [] [] [] []
//1 [] [] [] []
//2 [] [] [] []
//3 [] [] [] []
// . . . .
// . . . .
// . . . .
// . . . .
char[][] emptyBoard = new char[size][size];
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
emptyBoard[i][j] = '.';
}
}
Set<Integer> cols = new HashSet<Integer>();
Set<Integer> diagnoals = new HashSet<Integer>();
Set<Integer> antiDiagnoals = new HashSet<Integer>();
helper(0, cols, diagnoals, antiDiagnoals, emptyBoard, size, solutions);
return solutions;
}
private static void helper(int row, Set<Integer> cols, Set<Integer> diagnoals, Set<Integer> antiDiagnoals, char[][] board, int size, List<List<String>> solutions){
// base case
/* we can not use HashSet<Integer> cols, HashSet<Integer> diagnoals... ArrayList<List<String>> solutions;
it will happen: Set<Integer> cannot be converted to HashSet<Integer>
Not every List<Integer> is a ArrayList<Integer>. The dsp method accepts a ArrayList<Integer> so to make it work you'll pass a type ArrayList<Integer> or change your method to accept List<Integer>. Also, make use of generics, I'd use List<Integer> for the return type instead of List.
https://stackoverflow.com/questions/51430241/java-compile-error-listinteger-cannot-be-converted-to-arraylistinteger
*/
if (row == size){
solutions.add(transfer(board, size));
}
// recursion rule
for (int col = 0; col < size; col++){
int currDiagnoal = row - col;
int currAntiDiagnoal = row + col;
if (cols.contains(col) || diagnoals.contains(currDiagnoal) || antiDiagnoals.contains(currAntiDiagnoal)){
continue;
}
// eat
cols.add(col);
diagnoals.add(currDiagnoal);
antiDiagnoals.add(currAntiDiagnoal);
board[row][col] = 'Q';
helper(row + 1, cols, diagnoals, antiDiagnoals, board, size, solutions);
// spit
cols.remove(col);
diagnoals.remove(currDiagnoal);
antiDiagnoals.remove(currAntiDiagnoal);
board[row][col] = '.';
}
}
private static List<String> transfer(char[][] board, int size){
List<String> subSolutions = new ArrayList<String>();
for (int i = 0; i < size; i++){
String row = new String(board[i]);
subSolutions.add(row);
}
return subSolutions;
}
}
class Solution {
private int size; // 全局变量
private List<List<String>> solutions = new ArrayList<List<String>>();
public List<List<String>> solveNQueens(int n) {
size = n;
char emptyBoard[][] = new char[size][size];
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
emptyBoard[i][j] = '.';
}
}
helper(0, new HashSet<>(), new HashSet<>(), new HashSet<>(), emptyBoard);
return solutions;
}
private void helper(int row, Set<Integer> diagonals, Set<Integer> antiDiagonals, Set<Integer> cols, char[][] state){ // size 为全局变量, 不能用static访问
// Base case - N queens have been placed
if (row == size){
solutions.add(createBoard(state));
return;
}
for (int col = 0; col < size; col++){
int currDiagonal = row - col;
int currAntiDiagonal = row + col;
// If the queen is not placeable
if (cols.contains(col) || diagonals.contains(currDiagonal) || antiDiagonals.contains(currAntiDiagonal)) {
continue;
}
// "Add" the queen to the board 吃
cols.add(col);
diagonals.add(currDiagonal);
antiDiagonals.add(currAntiDiagonal);
state[row][col] = 'Q';
// Move on to the next row with the updated board state
helper(row + 1, diagonals, antiDiagonals, cols, state);
// "Remove" the queen from the board since we have already 吐
// explored all valid paths using the above function call
cols.remove(col);
diagonals.remove(currDiagonal);
antiDiagonals.remove(currAntiDiagonal);
state[row][col] = '.';
}
}
// Making use of a helper function to get the
// solutions in the correct output format
private List<String> createBoard(char[][] state) {
List<String> board = new ArrayList<String>();
for (int row = 0; row < size; row++){
String current_row = new String(state[row]);
board.add(current_row);
}
return board;
}
}
The diagonals are a litter trickier - but they have a property that we can use to our advantage.
- For each square on a given diagonal, the difference between the row and column indices
row-colwill be constant. Think about the diagonal that starts from(0,0)- the \(i^{th}\) square has the coordinates(i, i), so the difference is always 0.
- For each square on a given anti-diagonal, the sum of the row and column indexes
row + colwill be constant. If you were to start at the highest square in an anti-diagonal and more downwards, the row index increments by 1row + 1, and the column index decrements by 1(col -1). These cancel each other out.
Treating it looks like a math role, understanding it and then remember it. Using it directly.