Java-栈方法实现迷宫问题

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import java.util.Stack;public class Solution {    public final int M = 6;    public final int N = 8;    //8*10  M+2,N+2    public int[][] maze = {            {1, 1, 1, 1, 1, 1, 1, 1, 1, 1},            {1, 0, 1, 1, 1, 0, 1, 1, 1, 1},            {1, 1, 0, 1, 0, 1, 1, 1, 1, 1},            {1, 0, 1, 0, 0, 0, 0, 0, 1, 1},            {1, 0, 1, 1, 1, 0, 1, 1, 1, 1},            {1, 1, 0, 0, 1, 1, 0, 0, 0, 1},            {1, 0, 1, 1, 0, 0, 1, 1, 0, 1},            {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}    };    //8个移动方向的增量值,顺时针,东、东南、南......    public Direction directions[] = {            new Direction(0, 1),    //东            new Direction(1, 1),            new Direction(1, 0),            new Direction(1, -1),            new Direction(0, -1),            new Direction(-1, -1),            new Direction(-1, 0),            new Direction(-1, 1)    };    public boolean isExit() {        Stack s = new Stack();        int x, y, d;        int i, j;        //入口进站        s.push( new Item(1, 1, 0));        while (!s.empty()) {            Item item = (Item) s.pop();            x = item.x;            y = item.y;            d = item.d;            while (d < 8) {                i = x + directions[d].x;                j = y + directions[d].y;                if (maze[i][j] == 0) {                    s.push(new Item(x, y, d));                    x = i;                    y = j;                    maze[x][y] = 3;//标记,已经走过                    if (x == M && y == N) return true;                    else {                        d = 0;                    }                } else {                    d++;                }            }        }        return false;    }}class Direction {    int x, y;    public Direction(int x, int y) {        this.x = x;        this.y = y;    }}class Item {    int x, y, d;    public Item(int x, int y, int d) {        this.x = x;        this.y = y;        this.d = d;    }}

先贴代码,懒得上思路了,简单分析一下:

迷宫m行n列maze[m][n],0表示通,1表示不通。从一个点可以向8个方向探路,为了处理边角情况,迷宫的四周全部加一行一列,最后为maze[m+2][n+2]。

思路:

栈初始化,

将入口点push入栈,

while(栈不为空)

{

栈顶元素出站;

while(方向<8){

如果可走,入栈,求新点坐标,并赋值给当前点,如果等于出口,返回成果。否则方向=0;

否则,方向++;


}

}


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