[英]Optimizing a 2D array pathfinding algorithm
我試圖找到通過 map 的最短路徑,表示為 0 和 1 的矩陣,其中 0 是可通過的空間,1 是不可通過的牆壁。 入口在左上角 (0, 0),出口在右下角 (width - 1, height - 1)。
我的 function solution(int[][] map)
找到從入口到出口的最短路徑的長度,您可以在該路徑中移除一堵牆作為路徑的一部分。 路徑長度是您通過的節點總數,包括入口節點和出口節點。 map
的條件包括起始位置和結束位置始終可以通過 (0),map 將始終可以解決,盡管您可能需要也可能不需要移除牆壁,並且 Z1D78DC8ED51 的高度和寬度可以從 2 到 2214E51AEFEZB5114 20. 只能在大方向上移動; 不允許對角線移動。
幾個測試用例:
輸入:
int[][] a = {{0, 1, 1, 0},
{0, 0, 0, 1},
{1, 1, 0, 0},
{1, 1, 1, 0},
{1, 1, 1, 0}};
Output:
8
輸入:
int[][] b = {{0, 1, 0, 0, 0},
{0, 0, 0, 1, 0},
{0, 0, 1, 1, 0},
{0, 1, 1, 0, 0},
{0, 1, 1, 0, 0}};
Output:
9
輸入:
int[][] c = {{0, 1, 0, 0, 0},
{0, 0, 0, 1, 0},
{0, 0, 1, 1, 1},
{0, 1, 1, 0, 0},
{0, 1, 1, 0, 0}};
Output:
11
我的解決方案代碼如下:
public static int solution(int[][] map) {
int rows = map.length;
int cols = map[0].length;
Graph graph = new Graph(map);
Queue<Node> queue = new LinkedList<>();
ArrayList<Node> marked = new ArrayList<>();
Node start = graph.getNode(0, 0);
queue.add(start);
marked.add(start);
start.setDistanceFromStart(1);
while(!queue.isEmpty()) {
Node current = queue.remove();
if(current.getX() == rows - 1 && current.getY() == cols - 1) { //if we have reached goal node
return current.getDistanceFromStart();
}
for(Node n : current.getNeighbors()) {
queue.add(n);
n.setDistanceFromStart(current.getDistanceFromStart() + 1);
}
}
return -1; //no path found
}
我創建了兩個類以及解決方法,Graph 和 Node:
class Graph{
private Node[][] _nodeMap;
private int _rows;
private int _cols;
public Graph(int[][] map) {
_nodeMap = new Node[map.length][map[0].length];
_rows = _nodeMap.length;
_cols = _nodeMap[0].length;
for (int i = 0; i < _rows; i++) {
for(int j = 0; j < _cols; j++) {
_nodeMap[i][j] = new Node(i, j, map[i][j], false, this);
}
}
}
/**
* Gets the Node at location (x,y)
* @param x - the x val of the Node being retrieved
* @param y - the y val of the Node being retrieved
* @return
*/
public Node getNode(int x, int y) {
return _nodeMap[x][y];
}
/**
* Replace the node at x,y with the new node n.
* @param x
* @param y
* @param n
*/
public void setNode(int x, int y, Node n) {
_nodeMap[x][y] = n;
}
/**
* Checks to see if a node has a neighbor (either a wall or a path) to the north
* @param n - the node being checked
* @return boolean - true if there is a neighbor, false if there is not
*/
public boolean hasNorth(Node n) {
if(n.getX() > 0) {
return true;
} else {
return false;
}
}
/**
* Return's the neighbor to the north
* @param n - the current node
* @return Returns the Node to the north of n
*/
public Node getNorth(Node n) {
return _nodeMap[n.getX() - 1][n.getY()];
}
/**
* Checks to see if a node has a neighbor (either a wall or a path) to the south
* @param n - the node being checked
* @return boolean - true if there is a neighbor, false if there is not
*/
public boolean hasSouth(Node n) {
if(n.getX() < _rows - 1) {
return true;
} else {
return false;
}
}
/**
* Return's the neighbor to the south
* @param n - the current node
* @return Returns the Node to the south of n
*/
public Node getSouth(Node n) {
return _nodeMap[n.getX() + 1][n.getY()];
}
/**
* Checks to see if a node has a neighbor (either a wall or a path) to the east
* @param n - the node being checked
* @return boolean - true if there is a neighbor, false if there is not
*/
public boolean hasEast(Node n) {
if(n.getY() < _cols - 1) {
return true;
} else {
return false;
}
}
/**
* Return's the neighbor to the east
* @param n - the current node
* @return Returns the Node to the east of n
*/
public Node getEast(Node n) {
return _nodeMap[n.getX()][n.getY() + 1];
}
/**
* Checks to see if a node has a neighbor (either a wall or a path) to the west
* @param n - the node being checked
* @return boolean - true if there is a neighbor, false if there is not
*/
public boolean hasWest(Node n) {
if(n.getY() > 0) {
return true;
} else {
return false;
}
}
/**
* Return's the neighbor to the west
* @param n - the current node
* @return Returns the Node to the west of n
*/
public Node getWest(Node n) {
return _nodeMap[n.getX()][n.getY() - 1];
}
}
class Node {
private int _x; //x location
private int _y; //y location
private int _type; //1 if a wall, 0 if a path
private Graph _map;
private int _distFromStart;
private boolean _wallRemoved;
public Node(int x, int y, int type, boolean wallRemoved, Graph map){
_x = x;
_y = y;
_type = type;
_wallRemoved = wallRemoved;
_map = map;
_distFromStart = -1;
}
public int getX() {
return _x;
}
public int getY() {
return _y;
}
public int getType() {
return _type;
}
public boolean getWallRemoved() {
return _wallRemoved;
}
public int getDistanceFromStart() {
return _distFromStart;
}
public void setDistanceFromStart(int distance) {
_distFromStart = distance;
}
/**
* Returns an ArrayList<Node> containing the neighbors of a node.
* @return
*/
public ArrayList<Node> getNeighbors(){
ArrayList<Node> neighbors = new ArrayList<>();
if(this._wallRemoved) { //if a wall has already been removed
if(_map.hasWest(this) && _map.getWest(this)._type == 0) { //check west neighbor
Node node = _map.getWest(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), true, _map/*, node._timesEvaluated + 1*/);
neighbors.add(n);
}
if(_map.hasEast(this) && _map.getEast(this)._type == 0) { //check east neighbor
Node node = _map.getEast(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), true, _map);
neighbors.add(n);
}
if(_map.hasNorth(this) && _map.getNorth(this)._type == 0) { //check north neighbor
Node node = _map.getNorth(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), true, _map);
neighbors.add(n);
}
if(_map.hasSouth(this) && _map.getSouth(this)._type == 0) { //check south neighbor
Node node = _map.getSouth(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), true, _map/);
neighbors.add(n);
}
} else { //if a wall hasn't been removed yet
if(_map.hasWest(this)) { //check west neighbor
if(_map.getWest(this)._type == 1) { //if west neighbor is a wall
if(!this._wallRemoved) {
Node node = _map.getWest(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), true, _map);
neighbors.add(n);
}
} else { //if west neighbor is a path
Node node = _map.getWest(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), false, _map);
neighbors.add(n);
}
}
if(_map.hasEast(this)) { //check east neighbor
if(_map.getEast(this)._type == 1) {
if(!this._wallRemoved) {
Node node = _map.getEast(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), true, _map);
neighbors.add(n);
}
} else {
Node node = _map.getEast(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), false, _map);
neighbors.add(n);
}
}
if(_map.hasNorth(this)) { //check north neighbor
if(_map.getNorth(this)._type == 1) {
if(!this._wallRemoved) {
Node node = _map.getNorth(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), true, _map);
neighbors.add(n);
}
} else {
Node node = _map.getNorth(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), false, _map);
neighbors.add(n);
}
}
if(_map.hasSouth(this)) { //check south neighbor
if(_map.getSouth(this)._type == 1) {
if(!this._wallRemoved) {
Node node = _map.getSouth(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), true, _map);
neighbors.add(n);
}
} else {
Node node = _map.getSouth(this);
Node n = new Node(node.getX(), node.getY(), node.getType(), false, _map);
neighbors.add(n);
}
}
}
return neighbors;
}
}
我的代碼有效,為每個測試用例返回正確答案。 我的問題是代碼非常慢。 我很確定迷宮是通過蠻力解決的(如果不是,時間限制至少類似於蠻力)導致大型地圖(例如 20x20 地圖)需要很長時間才能解決。 我怎樣才能優化我的代碼以使其運行得更快? 謝謝!
你有這個 ArrayList 稱為“標記”。 我建議使用它;)
for(Node n : current.getNeighbors()) {
if(!marked.contains(n)){
queue.add(n);
n.setDistanceFromStart(current.getDistanceFromStart() + 1);
marked.add(n);
}
}
這將復雜度降低到 O(n * m),其中 n,m 是網格的尺寸。
另請閱讀關於 2d 空間中的 BFS 算法。 祝你好運:)
編輯#1
如果您想進一步改進代碼,請嘗試檢查A* algorithm 。
此外,您可以使用特殊參數將所有“進行中”節點保留在 PriorityQueue 上,而不是隊列,該參數將嘗試選擇最佳節點。 該參數可以是節點和(寬度 - 1,高度 - 1)(簡單畢達哥拉斯定理)之間的笛卡爾平面中的距離。 再次祝你好運:)
我認為這對普林斯頓大學的聯合查找會有所幫助https://www.cs.princeton.edu/~rs/AlgsDS07/01UnionFind.pdf
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