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++;
}
}