这是我在 java 中解决 n-queens 问题的代码。但是，当它应该是 92 时，输出是 0（在这种情况下是 8 个皇后的解数）。我们应该只使用堆栈和回溯（没有递归！！）。我真的被困住了！任何帮助将不胜感激！
import java.util.Stack;
public class NQueens {
//***** fill in your code here *****
//feel free to add additional methods as necessary
public static Stack<Integer> s = new Stack<Integer>();
public static int n ;
public static int total;
public static int i;
public static int Q;
//finds and prints out all solutions to the n-queens problem
public static int solve(int n) {
// i goes through each row to place a queen
// x goes through the columns within each row
for(int i = 0; i <n; i++) {
for(int x = 0; x<n; x++){
if(conflict(x) == false){ // loop through each column and checks whether it conflicts with current position of queen
s.push(x); // no conflict, push x
Q = s.get(x); // set current position of queen
break; //break out of loop to move on to next row, i++
}
else if (conflict(x)==true){
if(s.isEmpty() == true){
break;
}
if(x==n-1){ // if its looped through all columns, and there's no valid position
s.pop(); //pop last entry
i= -1; // and backtrack to previous row, to find another valid position for q in previous row
}
}
if (s.size()==n){ // if stack size is n, then stack is full and a solution has been found
total++;
System.out.print(s);// print solution
s.pop();
i= - 1; //backtrack to find next solution
}
}
}
return total;
}
public static boolean conflict(int k) {
if (Q==k|| (k-Q)== (i-(i-1))|| (Q-k)== (i-(i-1)) || k == s.pop()) {
return false; //there is a conflict!! k
}
return true; //is conflict
}
//this method prints out a solution from the current stack
//(you should not need to modify this method)
private static void printSolution(Stack<Integer> s) {
for (int i = 0; i < s.size(); i ++) {
for (int j = 0; j < s.size(); j ++) {
if (j == s.get(i))
System.out.print("Q ");
else
System.out.print("* ");
}//for
System.out.println();
}//for
System.out.println();
}//printSolution()
// ----- the main method -----
// (you shouldn't need to change this method)
public static void main(String[] args) {
int n = 8;
// pass in parameter n from command line
if (args.length == 1) {
n = Integer.parseInt(args[0].trim());
if (n < 1) {
System.out.println("Incorrect parameter");
System.exit(-1);
}//if
}//if
int number = solve(n);
System.out.println("There are " + number + " solutions to the " + n + "-queens problem.");
}//main()
}