使用递归回溯算法解决八皇后问题

什么是八皇后问题

八皇后问题，是一个古老而著名的问题，是回溯算法的典型案例。该问题是国际西洋棋棋手马克斯·贝瑟尔于1848年提出：在8×8格的国际象棋上摆放八个皇后，使其不能互相攻击，即任意两个皇后都不能处于同一行、同一列或同一斜线上，问有多少种摆法。

代码实现

package com.swh.data.recursion;

import com.alibaba.fastjson.JSON;

public class Queue8 {
    /**
     *  表示存放的皇后  数组元素的下标表示存放的第几个皇后也是第几行皇后
     *  数组元素中的值表示 皇后存放的第几列
     */
    int max = 8;
    private int[] arrays = new int[max];
    static int count = 0;

    public static void main(String[] args) {
        Queue8 queue8 = new Queue8();
        queue8.check(0);
        System.out.printf("一共有%d种排列方法",count);
    }

    /**
     * n  表示第几个房后
     *
     * @param n
     */
    private void check(int n) {
        if (n == max) { //n == 8 表示已经开始放第九个皇后(n 从0开始)  说明前八个已经放好  也就是说当前这种放置方式已经完成
            print();
            return;
        }

        for (int i = 0; i < max; i++) {
            arrays[n] = i; // 把当前的列设置到数组中
            // 检查放入的皇后是否与其他皇后冲突
            if(!judge(n)){
                check(n+1);
            }
        }

    }

    /**
     *  n 表示第几个皇后
     *  该方法用于判断当前第n个皇后 是否与前面几个皇后有冲突
     *
     * @param n
     * @return
     */
    private boolean judge(int n) {

        for (int i=0;i

执行结果

[0,4,7,5,2,6,1,3]
[0,5,7,2,6,3,1,4]
[0,6,3,5,7,1,4,2]
[0,6,4,7,1,3,5,2]
[1,3,5,7,2,0,6,4]
[1,4,6,0,2,7,5,3]
[1,4,6,3,0,7,5,2]
[1,5,0,6,3,7,2,4]
[1,5,7,2,0,3,6,4]
[1,6,2,5,7,4,0,3]
[1,6,4,7,0,3,5,2]
[1,7,5,0,2,4,6,3]
[2,0,6,4,7,1,3,5]
[2,4,1,7,0,6,3,5]
[2,4,1,7,5,3,6,0]
[2,4,6,0,3,1,7,5]
[2,4,7,3,0,6,1,5]
[2,5,1,4,7,0,6,3]
[2,5,1,6,0,3,7,4]
[2,5,1,6,4,0,7,3]
[2,5,3,0,7,4,6,1]
[2,5,3,1,7,4,6,0]
[2,5,7,0,3,6,4,1]
[2,5,7,0,4,6,1,3]
[2,5,7,1,3,0,6,4]
[2,6,1,7,4,0,3,5]
[2,6,1,7,5,3,0,4]
[2,7,3,6,0,5,1,4]
[3,0,4,7,1,6,2,5]
[3,0,4,7,5,2,6,1]
[3,1,4,7,5,0,2,6]
[3,1,6,2,5,7,0,4]
[3,1,6,2,5,7,4,0]
[3,1,6,4,0,7,5,2]
[3,1,7,4,6,0,2,5]
[3,1,7,5,0,2,4,6]
[3,5,0,4,1,7,2,6]
[3,5,7,1,6,0,2,4]
[3,5,7,2,0,6,4,1]
[3,6,0,7,4,1,5,2]
[3,6,2,7,1,4,0,5]
[3,6,4,1,5,0,2,7]
[3,6,4,2,0,5,7,1]
[3,7,0,2,5,1,6,4]
[3,7,0,4,6,1,5,2]
[3,7,4,2,0,6,1,5]
[4,0,3,5,7,1,6,2]
[4,0,7,3,1,6,2,5]
[4,0,7,5,2,6,1,3]
[4,1,3,5,7,2,0,6]
[4,1,3,6,2,7,5,0]
[4,1,5,0,6,3,7,2]
[4,1,7,0,3,6,2,5]
[4,2,0,5,7,1,3,6]
[4,2,0,6,1,7,5,3]
[4,2,7,3,6,0,5,1]
[4,6,0,2,7,5,3,1]
[4,6,0,3,1,7,5,2]
[4,6,1,3,7,0,2,5]
[4,6,1,5,2,0,3,7]
[4,6,1,5,2,0,7,3]
[4,6,3,0,2,7,5,1]
[4,7,3,0,2,5,1,6]
[4,7,3,0,6,1,5,2]
[5,0,4,1,7,2,6,3]
[5,1,6,0,2,4,7,3]
[5,1,6,0,3,7,4,2]
[5,2,0,6,4,7,1,3]
[5,2,0,7,3,1,6,4]
[5,2,0,7,4,1,3,6]
[5,2,4,6,0,3,1,7]
[5,2,4,7,0,3,1,6]
[5,2,6,1,3,7,0,4]
[5,2,6,1,7,4,0,3]
[5,2,6,3,0,7,1,4]
[5,3,0,4,7,1,6,2]
[5,3,1,7,4,6,0,2]
[5,3,6,0,2,4,1,7]
[5,3,6,0,7,1,4,2]
[5,7,1,3,0,6,4,2]
[6,0,2,7,5,3,1,4]
[6,1,3,0,7,4,2,5]
[6,1,5,2,0,3,7,4]
[6,2,0,5,7,4,1,3]
[6,2,7,1,4,0,5,3]
[6,3,1,4,7,0,2,5]
[6,3,1,7,5,0,2,4]
[6,4,2,0,5,7,1,3]
[7,1,3,0,6,4,2,5]
[7,1,4,2,0,6,3,5]
[7,2,0,5,1,4,6,3]
[7,3,0,2,5,1,6,4]
一共有92种排列方法