迷宫求解算法一直是算法学习的经典，实现自然也是多种多样，包括动态规划，递归等实现，这里我们使用穷举求解，加深对栈的理解和应用
定义Position类用于存储坐标点
起点坐标为（1，1），终点坐标为（8，8）
地图打印在最下面
class Position {
private int px;
private int py;
public Position(int px, int py) {
this.px = px;
this.py = py;
}
public int getPx() {
return px;
}
public void setPx(int px) {
this.px = px;
}
public int getPy() {
return py;
}
public void setPy(int py) {
this.py = py;
}
}这里我们简单介绍下move()函数
move函数分别向四个方向移动，然后将可行的path入栈.注意，这里栈元素中每个栈元素Position都是new出来的，栈中存的是reference，
注意看下面这种写法：
currentPosition.setPy(currentPosition.getPy()+1);
stacks.push(currentPosition);这种写法一度让我陷入困惑，因为pop出来的Position都是一样的，原因大家可能应该明白了。。。
public void move() {
if (moveRight()) {
Position temp = new Position(currentPosition.getPx() + 1, currentPosition.getPy());
test.add(temp);
stacks.push(temp);
} else if (moveBottom()) {
Position temp = new Position(currentPosition.getPx(), currentPosition.getPy() + 1);
test.add(temp);
stacks.push(temp);
} else if (moveTop()) {
Position temp = new Position(currentPosition.getPx(), currentPosition.getPy() - 1);
test.add(temp);
stacks.push(temp);
} else if (moveLeft()) {
Position temp = new Position(currentPosition.getPx() - 1, currentPosition.getPy());
test.add(temp);
stacks.push(temp);
} else {
currentPosition = stacks.pop();//若当前位置四个方向都走不通，则将当前位置出栈，继续遍历上一节点
}
}整体代码
class Position {
private int px;
private int py;
public Position(int px, int py) {
this.px = px;
this.py = py;
}
public int getPx() {
return px;
}
public void setPx(int px) {
this.px = px;
}
public int getPy() {
return py;
}
public void setPy(int py) {
this.py = py;
}
}
public class Maze {
private final Position start;//迷宫的起点final
private final Position end;//迷宫的终点final
private ArrayList<String> footPrint;//足迹
private ArrayList<Position> test;
private MyStack<Position> stacks;//自定义栈（也可以用java.util中的Stack栈）若想了解MyStack的实现，可以参考我的另一篇博客
private Position currentPosition;//定义当前位置
public Maze() {//集合，栈的初始化工作
start = new Position(1, 1);
end = new Position(8, 8);
currentPosition = start;
stacks = new MyStack<>();
test = new ArrayList<>();
}
public static final int map[][] = //定义地图10*10的方格
{{1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 0, 0, 1, 0, 0, 0, 1, 0, 1},
{1, 0, 0, 1, 0, 0, 0, 1, 0, 1},
{1, 0, 0, 0, 0, 1, 1, 0, 0, 1},
{1, 0, 1, 1, 1, 0, 0, 0, 0, 1},
{1, 0, 0, 0, 1, 0, 0, 0, 0, 1},
{1, 0, 1, 0, 0, 0, 1, 0, 0, 1},
{1, 0, 1, 1, 1, 0, 1, 1, 0, 1},
{1, 1, 0, 0, 0, 0, 0, 0, 0, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1}};
public static void printMap() {//打印地图
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (map[i][j] == 1) System.out.print(" ■");
else System.out.print(" ");
}
System.out.println();
}
}
public boolean moveTop() {//上移
String s = currentPosition.getPx() + "" + (currentPosition.getPy() - 1);
if ((map[currentPosition.getPx()][currentPosition.getPy() - 1] != 1) & !isArrived(s)) {
footPrint.add(s);
return true;
}
return false;
}
public boolean moveRight() {//右移
String s = (currentPosition.getPx() + 1) + "" + currentPosition.getPy();
if (map[currentPosition.getPx() + 1][currentPosition.getPy()] != 1 & !isArrived(s)) {
footPrint.add(s);
return true;
}
return false;
}
public boolean moveBottom() {//下移
String s = currentPosition.getPx() + "" + (currentPosition.getPy() + 1);
if ((map[currentPosition.getPx()][currentPosition.getPy() + 1] != 1) & !isArrived(s)) {
footPrint.add(s);
return true;
}
return false;
}
public boolean moveLeft() {//左移
String s = (currentPosition.getPx() - 1) + "" + currentPosition.getPy();
if ((map[currentPosition.getPx() - 1][currentPosition.getPy()] != 1) & !isArrived(s)) {
footPrint.add(s);
return true;
}
return false;
}
public boolean isArrived(String position) {//判断当前位置是否已经到打过
return footPrint.contains(position);
}
public void move() {//move函数分别向四个方向移动，然后将可行的path入栈
if (moveRight()) {
Position temp = new Position(currentPosition.getPx() + 1, currentPosition.getPy());
test.add(temp);
stacks.push(temp);
} else if (moveBottom()) {
Position temp = new Position(currentPosition.getPx(), currentPosition.getPy() + 1);
test.add(temp);
stacks.push(temp);
} else if (moveTop()) {
Position temp = new Position(currentPosition.getPx(), currentPosition.getPy() - 1);
test.add(temp);
stacks.push(temp);
} else if (moveLeft()) {
Position temp = new Position(currentPosition.getPx() - 1, currentPosition.getPy());
test.add(temp);
stacks.push(temp);
} else {
currentPosition = stacks.pop();//若当前位置四个方向都走不通，则将当前位置出栈，继续遍历上一节点
}
}
public static void main(String[] args) {
Maze m = new Maze();
m.footPrint = new ArrayList<>();
m.footPrint.add("11");
m.stacks.push(m.start);
while (m.currentPosition.getPx() != 8 || m.currentPosition.getPy() != 8) {
m.move();
}
printMap();
System.out.println("下面是足迹,长度是：" + m.footPrint.size());
m.printFootPrint();
}
public void printFootPrint() {
for (int i = 0; i < footPrint.size(); i++) {
System.out.print(footPrint.get(i) + ",");
}
System.out.println();
}
}大家可能会疑惑，为什么足迹是不连续的（例如：21，12）两个位置是走不通的，是因为在path遍历过程中存在跳栈，既当前位置走不通便会将当前位置的Position出栈（stacks.pop），然后继续上一节点遍历。
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