题目分析:
Move 会有以下几种情况：
- out of boarder =>
return -1- hit self =>
return -1
special case: 如果下一个点是 snake 的 tail, 是不会碰撞的。- eat food =>
return snake.size() - 1- just move by q =>
return snake.size() - 1
注意逻辑顺讯：1. remove and unmark tail first; 2. add and mark new head.
old tail 和 new head 可能是一个点。
Initial Solution
class SnakeGame {
Point[][] board;
LinkedList<Point> snake;
LinkedList<Point> foodList;
public SnakeGame(int width, int height, int[][] food) {
board = new Point[height][width];
for (int i = 0; i < height; i++) {
for (int j = 0; j < width; j++) {
board[i][j] = new Point(i, j);
}
}
snake = new LinkedList<>();
board[0][0].isOccupied = true;
snake.add(board[0][0]);
foodList = new LinkedList<>();
for (int[] position: food) {
Point p = board[position[0]][position[1]];
foodList.add(p);
}
}
public int move(String direction) {
Point head = snake.getFirst();
int nextX = head.x;
int nextY = head.y;
switch (direction) {
case "U":
nextX = head.x - 1;
break;
case "L":
nextY = head.y - 1;
break;
case "R":
nextY = head.y + 1;
break;
case "D":
nextX = head.x + 1;
break;
default:
return -1;
}
// out of boarder
if (nextX < 0 || nextY < 0 || nextX >= board.length || nextY >= board[0].length) {
return -1;
}
Point nextPoint = board[nextX][nextY];
// hit self
if (nextPoint.isOccupied && !nextPoint.equals(snake.getLast())) {
return -1;
}
// eat food
if (foodList.size() > 0 && nextPoint.equals(foodList.getFirst())) {
snake.add(0, foodList.removeFirst());
nextPoint.isOccupied = true;
return snake.size() - 1;
}
// remove tail first, because tail can be the next head position
Point tail = snake.removeLast();
tail.isOccupied = false;
snake.add(0, nextPoint);
nextPoint.isOccupied = true;
return snake.size() - 1;
}
}
class Point {
int x;
int y;
boolean isOccupied;
public Point(int x, int y) {
this.x = x;
this.y = y;
this.isOccupied = false;
}
}
TLE with case 1000 * 1000 board.
这里使用 Point isOccupied 来标记该 point 是否是 snake 的一部分，或者说已经被 snake 占用。也就不需要 HashSet 来就行标记。
这个 Test Case 中的 board 很大 1000 x1000，但只使用了很小一部分 3 x 3。因而 constructor 的时间复杂度为 weight x height，超出了test设置的时间限度。
下面的 Solution 是对上面 case 进行空间上的优化，只存使用过的 Point，而不对这个 board 进行 Point 初始化。但实际使用上来说，Initial Solution 还是更好一些，毕竟整个 board 对玩家来说都是可以使用的，在玩家游戏的过程中进行 Hash 运算，会慢一些。
Solution 2
tried to use LinkedHashSet, but LinkedHashSet does not support add(index, element), which is needed for the case of eating food. So used HashSet and LinkedList instead.
这里的 LinkedList<Point> 来表示 snake，实际上使用了 interface Deque. 头尾均可添加、删除。LinkedList implements Deque.
遇到知识盲点了，debug 了很长时间。对于 class Point, 要使用 HashSet 和 HashMap，需要 Override 两个 methods from class Object.
public boolean equals(Object o); // pay attention here
// WRONG: public boolean equals(Point p)
public int hashCode();
之前由于使用了 public boolean equals(Point p) 导致HashSet contains not working as expected:
// same point p1, p2
Point p1 = new Point(1, 2);
Point p2 = new Point(1, 2);
Set<Point> set = new HashSet<>();
set.add(p1);
set.contains(p2); // false
这也说明了使用 @Override 的重要性。
这里把纠正后的 class Point 写一下，下面的 Solution 都是用这个 implementation：
class Point {
int x;
int y;
public Point(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public boolean equals(Object o) {
if (!(o instanceof Point)) {
return false;
}
Point p = (Point)o;
return this.x == p.x && this.y == p.y;
}
// WRONG
// public boolean equals(Point p) {
// return this.x == p.x && this.y == p.y;
// }
@Override
public int hashCode() {
return (x + ";" + y).hashCode();
}
}
class SnakeGame {
int height, width;
LinkedList<Point> snake;
Set<Point> pointsInUse;
LinkedList<Point> foodList;
public SnakeGame(int width, int height, int[][] food) {
this.height = height;
this.width = width;
snake = new LinkedList<>();
pointsInUse = new HashSet<>();
Point head = new Point(0, 0);
snake.add(head);
pointsInUse.add(head);
foodList = new LinkedList<>();
for (int[] position: food) {
Point p = new Point(position[0], position[1]);
foodList.add(p);
}
}
public int move(String direction) {
Point head = snake.getFirst();
int nextX = head.x;
int nextY = head.y;
switch (direction) {
case "U":
nextX = head.x - 1;
break;
case "L":
nextY = head.y - 1;
break;
case "R":
nextY = head.y + 1;
break;
case "D":
nextX = head.x + 1;
break;
default:
return -1;
}
// out of boarder
if (nextX < 0 || nextY < 0 || nextX >= height || nextY >= width) {
return -1;
}
Point nextPoint = new Point(nextX, nextY);
// hit self
if (!nextPoint.equals(snake.getLast()) && pointsInUse.contains(nextPoint)) {
return -1;
}
// eat food
if (foodList.size() > 0 && nextPoint.equals(foodList.getFirst())) {
foodList.removeFirst();
snake.add(0, nextPoint);
pointsInUse.add(nextPoint);
return snake.size() - 1;
}
// remove tail first, because tail can be the next head position
Point tail = snake.removeLast();
pointsInUse.remove(tail);
snake.add(0, nextPoint);
pointsInUse.add(nextPoint);
return snake.size() - 1;
}
}
class Point {...}
Solution 3
从只是解题的角度来考虑，使用 x * height + y 来表达 Point 更为高效。Solution 1 & 2 are more Object Oriented!
part1
part2
网友评论