N皇后问题使用stack的非递归实现
package com.xx.dataStructure.stack;
class Point {
int x;
int y;
int value;
Point(int x ,int y){
this(x,y,0);
}
Point(int x ,int y,int value){
this.x = x;
this.y = y;
this.value = value;
}
@Override
public String toString(){
return "(" + x + "," + y + ")";
}
}
//求解N皇后问题的所有解
public class Queen {
private int solution = 0;
private int n;
private Point [][] points = null;
public Queen(int n) {
assert (n > 0);
this.n = n;
this.solution = 0;
points = new Point[n][n];
for(int i =0; i<n ; ++i){
for(int j =0; j<n ; ++j){
points[i][j] = new Point(i,j);
}
}
}
//point的行、列、对角线的点value +1
public static void add(Point[][] points,Point point){
//列
for(int i = 0; i < points.length ; i++ ){
points[i][point.y].value += 1;
}
//行
for(int i = 0; i < points.length ; i++ ){
points[point.x][i].value += 1;
}
//对角线
for( int i= point.x, j = point.y ;( i < points.length && j < points.length);
++i , ++j ){
points[i][j].value += 1;
}
for( int i= point.x, j = point.y ;( i >= 0 && j < points.length);
--i , ++j ){
points[i][j].value += 1;
}
for( int i= point.x, j = point.y ;( i < points.length && j >= 0);
++i , --j ){
points[i][j].value += 1;
}
for( int i= point.x, j = point.y ;( i >= 0 && j >= 0);
--i , --j ){
points[i][j].value += 1;
}
point.value = 1;
}
public static void minus(Point[][] points,Point point){
//列
for(int i = 0; i < points.length ; i++ ){
points[i][point.y].value -= 1;
}
//行
for(int i = 0; i < points.length ; i++ ){
points[point.x][i].value -= 1;
}
//对角线
for( int i= point.x, j = point.y ;( i < points.length && j < points.length);
++i , ++j ){
points[i][j].value -= 1;
}
for( int i= point.x, j = point.y ;( i >= 0 && j < points.length);
--i , ++j ){
points[i][j].value -= 1;
}
for( int i= point.x, j = point.y ;( i < points.length && j >= 0);
++i , --j ){
points[i][j].value -= 1;
}
for( int i= point.x, j = point.y ;( i >= 0 && j >= 0);
--i , --j ){
points[i][j].value -= 1;
}
point.value = 0;
}
// 能否使用非递归算法 回溯法来使时间复杂度降为多项式
public static void solveProblem0(Queen queen) {
Stack<Point> stack = new Stack<Point>(100);
System.out.println("开始求解"+queen.n+"皇后问题。");
Point [][] points = queen.points;
int [] tags = new int[queen.n];
for(int i = 0; i < queen.n; i ++){
tags[i] = -1;
}
int k = 0;
try {
for (int i = 0; i < queen.n; ++i) {
Point currentPoint = points[0][i];
stack.push(currentPoint);
add(points,currentPoint);
k++;
while(!stack.isEmpty()){
if (k == queen.n ){
queen.solution++;
stack.traverse();
Point point = stack.pop();
minus(points, point);
k--;
}else {
Point nextP = findNextPoint(points,tags,k);
if (nextP != null){
stack.push(nextP);
add(points,nextP);
k++;
}else {
Point point = stack.pop();
minus(points, point);
tags[k--] = -1;
}
}
}
}
} catch (Exception e) {
e.printStackTrace();
}
}
public static Point findNextPoint(Point [][] points,int[] tags ,int k){
Point p = null;
for(int q = tags[k] + 1 ; q < points.length; q++){
tags[k]++;
if (points[k][q].value == 0){
p = points[k][q];
break;
}
}
return p;
}
public static void main(String[] args) {
solveProblem0(new Queen(1));
solveProblem0(new Queen(2));
solveProblem0(new Queen(3));
solveProblem0(new Queen(4));
solveProblem0(new Queen(5));
solveProblem0(new Queen(6));
}
}
程序输出
开始求解1皇后问题。 (0,0), 开始求解2皇后问题。 开始求解3皇后问题。 开始求解4皇后问题。 (0,1),(1,3),(2,0),(3,2), (0,2),(1,0),(2,3),(3,1), 开始求解5皇后问题。 (0,0),(1,2),(2,4),(3,1),(4,3), (0,0),(1,3),(2,1),(3,4),(4,2), (0,1),(1,3),(2,0),(3,2),(4,4), (0,1),(1,4),(2,2),(3,0),(4,3), (0,2),(1,0),(2,3),(3,1),(4,4), (0,2),(1,4),(2,1),(3,3),(4,0), (0,3),(1,0),(2,2),(3,4),(4,1), (0,3),(1,1),(2,4),(3,2),(4,0), (0,4),(1,1),(2,3),(3,0),(4,2), (0,4),(1,2),(2,0),(3,3),(4,1), 开始求解6皇后问题。 (0,1),(1,3),(2,5),(3,0),(4,2),(5,4), (0,2),(1,5),(2,1),(3,4),(4,0),(5,3), (0,3),(1,0),(2,4),(3,1),(4,5),(5,2), (0,4),(1,2),(2,0),(3,5),(4,3),(5,1),