Leetcode.52. N皇后问题

ゝ一纸荒年。 2022-12-15 12:59 158阅读 0赞

# 51.n-queens #

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

![在这里插入图片描述][20201017110926397.png_pic_center]  
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.

## 代码 ##

### 基于集合的回溯 ###

package n_queens;
    
    import java.util.ArrayList;
    import java.util.Arrays;
    import java.util.HashSet;
    import java.util.List;
    import java.util.Set;
    
    /** * n 皇后问题研究的是如何将 n 个皇后放置在 n×n 的棋盘上，并且使皇后彼此之间不能相互攻击 * 注意,1.每一行要都放一个皇后,棋盘是方形的 * */
    public class Solution1 { 
    	
        public List<List<String>> solveNQueens(int n) { 
            List<List<String>> solutions = new ArrayList<List<String>>();
            int[] queens = new int[n];//index表示第几行,index对应的数字表示第几列
            Arrays.fill(queens, -1);//填充为-1
            Set<Integer> columns = new HashSet<Integer>();//列集合
            Set<Integer> diagonals1 = new HashSet<Integer>();//斜线为从左上到右下方向
            Set<Integer> diagonals2 = new HashSet<Integer>();//斜线为从右上到左下方向
            backtrack(solutions, queens, n, 0, columns, diagonals1, diagonals2);
            return solutions;
        }
    
        /** * 基于集合的回溯,通过三个集合来判断是否包含 * @param solutions 结果集 * @param queens 填充数组,数组长度为n,下标表示行,元素值表示当前行的位置.有点一维数组当二维数组用的意思 * @param n n个皇后 * @param row 行 * @param columns 列集合 * @param diagonals1 斜线为从左上到右下方向 * @param diagonals2 斜线为从右上到左下方向 */
        public void backtrack(List<List<String>> solutions, int[] queens, int n, int row, 
            Set<Integer> columns, Set<Integer> diagonals1, Set<Integer> diagonals2) { 
            if (row == n) { 
                List<String> board = generateBoard(queens, n);//返回字符串
                solutions.add(board);
            } else { 
                for (int i = 0; i < n; i++) { //遍历当行行
                    if (columns.contains(i)) { 
                        continue;
                    }
                    //重点
                    int diagonal1 = row - i;//方向一的斜线为从左上到右下方向，同一条斜线上的每个位置满足行下标与列下标之差相等
                    if (diagonals1.contains(diagonal1)) { 
                        continue;
                    }
                    //重点
                    int diagonal2 = row + i;//方向二的斜线为从右上到左下方向，同一条斜线上的每个位置满足行下标与列下标之和相等
                    if (diagonals2.contains(diagonal2)) { 
                        continue;
                    }
                    queens[row] = i;
                    columns.add(i);//将当前列占用
                    diagonals1.add(diagonal1);//将当前斜线占用
                    diagonals2.add(diagonal2);//将当前斜线占用
                    //进入下一行
                    backtrack(solutions, queens, n, row + 1, columns, diagonals1, diagonals2);
                    //回溯
                    queens[row] = -1;
                    columns.remove(i);
                    diagonals1.remove(diagonal1);
                    diagonals2.remove(diagonal2);
                }
            }
        }
    
        /**数组转字符串*/
        public List<String> generateBoard(int[] queens, int n) { 
            List<String> board = new ArrayList<String>();
            for (int i = 0; i < n; i++) { 
                char[] row = new char[n];
                Arrays.fill(row, '.');
                row[queens[i]] = 'Q';
                board.add(new String(row));
            }
            return board;
        }
        
        public static void main(String[] args) { 
        	//测试4*4的棋盘
    		List<List<String>> list = new Solution1().solveNQueens(4);
    		list.forEach(list2->{ 
    			list2.forEach(System.out::println);
    			System.out.println("=====");
    		});
    	}
    }

### 方法二：基于位运算的回溯(不推荐) ###

![在这里插入图片描述][20201017111757585.png]  
我们可以用数组来代替集合。  
放完了第0行，第1行，那里第2行怎么放呢。  
我们用 0 代表可以放置皇后的位置，1 代表不能放置皇后的位置。  
很明显，情况如下。

列          ， 从左上到右下方向  ，从右上到左下方向
    0001 0100   ， 0011 0000       ，0000 1001

图片来自leetcode官方

class Solution { 
        public List<List<String>> solveNQueens(int n) { 
            int[] queens = new int[n];
            Arrays.fill(queens, -1);
            List<List<String>> solutions = new ArrayList<List<String>>();
            solve(solutions, queens, n, 0, 0, 0, 0);
            return solutions;
        }
    
        public void solve(List<List<String>> solutions, int[] queens, int n, int row, int columns, int diagonals1, int diagonals2) { 
            if (row == n) { 
                List<String> board = generateBoard(queens, n);
                solutions.add(board);
            } else { 
                int availablePositions = ((1 << n) - 1) & (~(columns | diagonals1 | diagonals2));
                while (availablePositions != 0) { 
                    int position = availablePositions & (-availablePositions);
                    availablePositions = availablePositions & (availablePositions - 1);
                    int column = Integer.bitCount(position - 1);
                    queens[row] = column;
                    solve(solutions, queens, n, row + 1, columns | position, (diagonals1 | position) << 1, (diagonals2 | position) >> 1);
                    queens[row] = -1;
                }
            }
        }
    
        public List<String> generateBoard(int[] queens, int n) { 
            List<String> board = new ArrayList<String>();
            for (int i = 0; i < n; i++) { 
                char[] row = new char[n];
                Arrays.fill(row, '.');
                row[queens[i]] = 'Q';
                board.add(new String(row));
            }
            return board;
        }
    }

# 52.n-queens-ii #

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

![在这里插入图片描述][20201017110926397.png_pic_center]

Given an integer n, return the number of distinct solutions to the n-queens puzzle.

Example:

Input: 4
    Output: 2
    Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
    [
     [".Q..",  // Solution 1
      "...Q",
      "Q...",
      "..Q."],
    
     ["..Q.",  // Solution 2
      "Q...",
      "...Q",
      ".Q.."]
    ]

# java代码 #

class Solution { 
        public int totalNQueens(int n) { 
            Set<Integer> columns = new HashSet<Integer>();
            Set<Integer> diagonals1 = new HashSet<Integer>();
            Set<Integer> diagonals2 = new HashSet<Integer>();
            return backtrack(n, 0, columns, diagonals1, diagonals2);
        }
    
        public int backtrack(int n, int row, Set<Integer> columns, Set<Integer> diagonals1, Set<Integer> diagonals2) { 
            if (row == n) { 
                return 1;
            } else { 
                int count = 0;
                for (int i = 0; i < n; i++) { 
                    if (columns.contains(i)) { 
                        continue;
                    }
                    int diagonal1 = row - i;
                    if (diagonals1.contains(diagonal1)) { 
                        continue;
                    }
                    int diagonal2 = row + i;
                    if (diagonals2.contains(diagonal2)) { 
                        continue;
                    }
                    columns.add(i);
                    diagonals1.add(diagonal1);
                    diagonals2.add(diagonal2);
                    count += backtrack(n, row + 1, columns, diagonals1, diagonals2);
                    columns.remove(i);
                    diagonals1.remove(diagonal1);
                    diagonals2.remove(diagonal2);
                }
                return count;
            }
        }
    }

或者

class Solution { 
        public int totalNQueens(int n) { 
            return solve(n, 0, 0, 0, 0);
        }
    
        public int solve(int n, int row, int columns, int diagonals1, int diagonals2) { 
            if (row == n) { 
                return 1;
            } else { 
                int count = 0;
                int availablePositions = ((1 << n) - 1) & (~(columns | diagonals1 | diagonals2));
                while (availablePositions != 0) { 
                    int position = availablePositions & (-availablePositions);
                    availablePositions = availablePositions & (availablePositions - 1);
                    count += solve(n, row + 1, 
                    columns | position, 
                    (diagonals1 | position) << 1, 
                    (diagonals2 | position) >> 1);
                }
                return count;
            }
        }
    }

或者

class Solution { 
        public int totalNQueens(int n) { 
                    int[] rs = new int[]{ 0,1,0,0,2,10,4,40,92,352,724,2680};
            return rs[n];
        }
    }

[20201017110926397.png_pic_center]: /images/20221123/a7e0efe3999f4c8fa4d42f6b4f6ec38a.png
[20201017111757585.png]: /images/20221123/fa0b38a5cf384188974955f1e11abf54.png