N皇后问题的Java程序|回溯-3

N皇后是在N×N棋盘上放置N个棋后的问题，这样没有两个皇后相互攻击。例如，以下是 4 Queen 问题的解决方案。

预期的输出是一个二进制矩阵，其中放置皇后的块为 1。例如，以下是上述 4 个皇后解决方案的输出矩阵。

{ 0,  1,  0,  0}
              { 0,  0,  0,  1}
              { 1,  0,  0,  0}
              { 0,  0,  1,  0}

Java

/* Java program to solve N Queen Problem using
   backtracking */
public class NQueenProblem {
    final int N = 4;
 
    /* A utility function to print solution */
    void printSolution(int board[][])
    {
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++)
                System.out.print(" " + board[i][j]
                                 + " ");
            System.out.println();
        }
    }
 
    /* A utility function to check if a queen can
       be placed on board[row][col]. Note that this
       function is called when "col" queens are already
       placed in columns from 0 to col -1. So we need
       to check only left side for attacking queens */
    boolean isSafe(int board[][], int row, int col)
    {
        int i, j;
 
        /* Check this row on left side */
        for (i = 0; i < col; i++)
            if (board[row][i] == 1)
                return false;
 
        /* Check upper diagonal on left side */
        for (i = row, j = col; i >= 0 && j >= 0; i--, j--)
            if (board[i][j] == 1)
                return false;
 
        /* Check lower diagonal on left side */
        for (i = row, j = col; j >= 0 && i < N; i++, j--)
            if (board[i][j] == 1)
                return false;
 
        return true;
    }
 
    /* A recursive utility function to solve N
       Queen problem */
    boolean solveNQUtil(int board[][], int col)
    {
        /* base case: If all queens are placed
           then return true */
        if (col >= N)
            return true;
 
        /* Consider this column and try placing
           this queen in all rows one by one */
        for (int i = 0; i < N; i++) {
            /* Check if the queen can be placed on
               board[i][col] */
            if (isSafe(board, i, col)) {
                /* Place this queen in board[i][col] */
                board[i][col] = 1;
 
                /* recur to place rest of the queens */
                if (solveNQUtil(board, col + 1) == true)
                    return true;
 
                /* If placing queen in board[i][col]
                   doesn't lead to a solution then
                   remove queen from board[i][col] */
                board[i][col] = 0; // BACKTRACK
            }
        }
 
        /* If the queen can not be placed in any row in
           this column col, then return false */
        return false;
    }
 
    /* This function solves the N Queen problem using
       Backtracking.  It mainly uses solveNQUtil () to
       solve the problem. It returns false if queens
       cannot be placed, otherwise, return true and
       prints placement of queens in the form of 1s.
       Please note that there may be more than one
       solutions, this function prints one of the
       feasible solutions.*/
    boolean solveNQ()
    {
        int board[][] = { { 0, 0, 0, 0 },
                          { 0, 0, 0, 0 },
                          { 0, 0, 0, 0 },
                          { 0, 0, 0, 0 } };
 
        if (solveNQUtil(board, 0) == false) {
            System.out.print("Solution does not exist");
            return false;
        }
 
        printSolution(board);
        return true;
    }
 
    // driver program to test above function
    public static void main(String args[])
    {
        NQueenProblem Queen = new NQueenProblem();
        Queen.solveNQ();
    }
}
// This code is contributed by Abhishek Shankhadhar

输出：

0  0  1  0 
1  0  0  0 
0  0  0  1 
0  1  0  0

请参阅有关 N 皇后问题的完整文章 | Backtracking-3 了解更多详情！