leetcode：N-Queens 【Java】

最新推荐文章于 2023-06-03 18:35:46 发布

原创最新推荐文章于 2023-06-03 18:35:46 发布 · 652 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

79 篇文章

订阅专栏

本文介绍了一种使用回溯算法解决N皇后问题的方法，并提供了一个详细的Java实现案例。该算法能够在N×N的棋盘上放置N个皇后，确保任意两个皇后都不会互相攻击。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

1、采用回溯算法；

2、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;
    }
}