1. An application of using Stacks
Going through a Maze
Try This
Principle: Remember the return path in case we have to go back
11/27/2018
IT 179

2. Play Maze public class Maze {
/**
* Test if p is a valid move, i.e., a space ' ' in m[p.x][p.y].
m char[][], a 2-dimensional char array.
p Point, a position */
private static boolean validMove(char[][] m, Point p) {
return p.x >= 0 && p.x < m.length &&
p.y >= 0 && p.y < m[0].length &&
m[p.x][p.y] == ' '; }
...
11/27/2018
IT 179

3. Play Maze public class Maze { ...
/**
* Try to find next move from current position */
static Point nextMove(char[][] m, Point current) {
Point next;
next = new Point(current.x,current.y+1); // try east
if (validMove(m,next)) return next;
next = new Point(current.x+1,current.y); // try south
next = new Point(current.x,current.y-1); // try west
next = new Point(current.x-1,current.y); // try north
return null; }
11/27/2018
IT 179

4. /* Solving a maze: put '.' to the exit path, 'X' a failed try. */
public static LStack<Point> solve(char[][] m, Point start, Point exit) {
LStack<Point> path = new LStack<Point>();
Point next, current = new Point(start);
while (current.x != exit.x || current.y != exit.y) {
next = nextMove(m, current); // try to find next move
if (next == null) { // no possible next move
if (path.empty()) // if empty, no way to return;
return null; // i.e., no solution
m[current.x][current.y] = 'X'; // don't come back again
current = path.pop();
} else {
m[current.x][current.y] = '.';
path.push(current);
current = next; }
return path;
11/27/2018
IT 179