java使用链栈实现迷宫求解
java实现链栈在前面有所介绍：http://www.cnblogs.com/lixiaolun/p/4644141.html
java实现链栈的代码：
package stackapplication;
public class LinkStack {
private Element base;
private Element top;
class Element
{
public Step data;
public Element next;
}
/**
* 初始化栈
* */
public void initStack()
{
top = new Element();
base = new Element();
top.data=null;
top.next=base;
base.data=null;
base.next=null;
}
/**
* 入栈
* */
public void push(Step o)
{
Element e = new Element();
e.data = o;
if(top.next==base)//第一次入栈操作
{
e.next=base;
top.next=e;
}else
{
e.next=top.next;
top.next=e;
}
}
/**
* 出栈
* */
public Step pop()
{
Step o = null;
if(top.next==base)
{
System.out.println("栈中没有元素！");
return o;
}else
{
o = top.next.data;
//System.out.println("出栈操作"+o);
top.next=top.next.next;
}
return o;
}
/**
* 判断栈是否为空
* */
public Boolean isEmpty()
{
if(top.next==base)
{
return true;
}
return false;
}
/**
* 打印栈
* */
public void print()
{
System.out.print("打印栈：");
Element temp =top;
while(temp.next!=base)
{
System.out.print(temp.next.data+"\t");
temp =temp.next;
}
System.out.println();
}
}
java实现迷宫求解的类代码：
package stackapplication;
public class Maze {
public static void main(String[] args) {
int [][]map={
{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}
};//入口在map[1][1],出口在map[8][8]
int [][]move={{0,-1},{0,1},{-1,0},{1,0}};//上下左右四个移动方向
LinkStack s = new LinkStack();
s.initStack();
LinkStack s1 = new LinkStack();
s1.initStack();
path(map,move,s,s1);
while(!s1.isEmpty())
{
Step step = s1.pop();
System.out.println("("+step.x+","+step.y+")");
}
}
private static int path(int[][] map, int[][] move,LinkStack s,LinkStack s1) {
Step step = new Step(1, 1, -1);//起始位置
//map[1][1]=-1;//表示已走过该点
s.push(step);
s1.push(step);
while(!s.isEmpty())
{
step=s.pop();
int x=step.x;
int y=step.y;
int d=step.d+1;
while(d<4)
{
int i=x+move[d][0];
int j=y+move[d][1];
if(map[i][j]==0 && i>=0 && i<10 && j>=0 &&j<10)//该位置是通的，且不越界
{
System.out.println(i+","+j);
step = new Step(x, y, d);
s.push(step);//将当前位置压入栈顶
s1.push(step);
step = new Step(i, j, d);
s.push(step);//将当前位置压入栈顶
s1.push(step);
x=i;
y=j;
map[x][y]=-1;//表示已走过该点
if(x==8 && y==8)//到达出口
{
System.out.println("到达出口");
return 1;
}else
{
d=0;//到达一个新的点，所以要从新初始化方向，遍历其4个方向是否是通的
}
}else
{
d++;//下一个方向
}
}
}
return 0;
}
}
class Step
{
int x;//横坐标
int y;//纵坐标
int d;//移动方向，取值为0,1,2,3。分别表示上下左右4个方向。
public Step(int x,int y,int d)
{
this.x=x;
this.y=y;
this.d=d;
}
}