一、问题描述
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.."] ]二、问题分析
找到一个可行解之后,还需要继续搜索。
三、算法代码
public class Solution { public List<List<String>> solveNQueens(int n) { List<List<String>> result = new ArrayList<>(); if (n <= 0) return result; char[][] board = new char[n][n]; for (char[] row : board) { Arrays.fill(row, '.'); } boolean[] col_occupied = new boolean[n]; placeQueen(result, board, col_occupied, 0, n); return result; } private void placeQueen(List<List<String>> result, char[][] board, boolean[] col_occupied, int rowNum, int n) { if (rowNum == n) { List<String> list = new ArrayList<String>(); for (char[] row : board) { list.add(new String(row)); } result.add(list); return; } for (int colNum=0; colNum<n; colNum++) { if (isValid(board, col_occupied, rowNum, colNum, n)){ board[rowNum][colNum] = 'Q'; col_occupied[colNum] = true; placeQueen(result, board, col_occupied, rowNum+1, n); board[rowNum][colNum] = '.'; //回溯,尝试皇后rowNum的下一个位置 col_occupied[colNum] = false; } } } private boolean isValid(char[][]board, boolean[] col_occupied, int row, int col, int n) { if (col_occupied[col]) return false; for (int i=1; row-i>=0 && col-i>=0; i++) { if (board[row-i][col-i] == 'Q') return false;//反对角斜线 } for (int i=1; row-i>=0 && col+i<n; i++) { if (board[row-i][col+i] == 'Q') return false;//正对角斜线 } return true; } }