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.."]
]
一刷
题解：
任两个皇后都不能处于同一条横行、纵行或斜线上。DFS求解，注意check board是否validate的条件：board[i][j] == 'Q' && (x + j == y + i || x + y == i + j || x == i)
public class Solution {
public List<List<String>> solveNQueens(int n) {
List<List<String>> res = new ArrayList<>();
char[][] board = new char[n][n];
for(int i=0; i<n; i++){
for(int j=0; j<n; j++){
board[i][j] = '.';
}
}
dfs(board, 0, res);
return res;
}
private void dfs(char[][] board, int colIndex, List<List<String>> res) {
if(colIndex == board.length) {
res.add(construct(board));
return;
}
for(int i = 0; i < board.length; i++) {
if(validate(board, i, colIndex)) {
board[i][colIndex] = 'Q';
dfs(board, colIndex + 1, res);
board[i][colIndex] = '.';
}
}
}
private boolean validate(char[][] board, int x, int y) {
for(int i = 0; i < board.length; i++) {
for(int j = 0; j < y; j++) {
if(board[i][j] == 'Q' && (x + j == y + i || x + y == i + j || x == i))//
return false;
}
}
return true;
}
private List<String> construct(char[][] board){
List<String> res = new LinkedList<String>();
for(int i=0; i<board.length; i++){
String s = new String(board[i]);
res.add(s);
}
return res;
}
}
