算法
详细讲解和注释请看
https://blog.csdn.net/m0_68055637/article/details/128960743
java 代码
import java.util.Scanner;
public class Main{
static int N=11,n;
static char [][]q=new char[N][N]; //存储棋盘
static boolean []cor=new boolean[N]; //判断列是否有皇后
static boolean []dg=new boolean[N*2]; //判断对角线是否有皇后,n * n的矩阵,存在r + i也就是行加上列求截距的操作,必须开两倍大否则就爆了
static boolean []udg=new boolean[N*2]; //判断反对角线是否有皇后
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
n=sc.nextInt();
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
q[i][j]='.';
}
}
dfs(0);
}
private static void dfs(int r) {
if(r==n){ //代表棋盘处理完毕,是结束出口
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
System.out.print(q[i][j]);
}
System.out.println();
}
System.out.println();
return;
}
for (int i = 0; i < n; i++) { //第r行,第i列是否可以放皇后
//对角线y1=x1+b1,y2=-x2+b2,那么b1=y1-x1,b2=y2+x2
//为了防止y1-x1是个负数,加上偏移量n
if(!cor[i] &&!dg[i+r] &&!udg[r-i+n]){
q[r][i]='Q';
cor[i]=dg[i+r]=udg[r-i+n]=true;
dfs(r+1); //去下一行遍历
cor[i]=dg[i+r]=udg[r-i+n]=false; //恢复现场
q[r][i]='.';
}
}
}
}