我大胆尝试学习并在游戏中实现这个算法，“理解”它相对容易。但在实现过程中，我一直遇到一个无法退出的无限循环。我反复检查了代码，但肯定遗漏了一些东西，或者没有理解某些部分。 任何帮助将不胜感激。

public class PathFinder {private LinkedList<Node> openList;private LinkedList<Node> closedList;public PathFinder() {    openList = new LinkedList<Node>();    closedList = new LinkedList<Node>();}private List<Node> buildPath(Node node) {    LinkedList<Node> path = new LinkedList<Node>();    while (node.parentNode != null) {        path.addFirst(node);        node = node.parentNode;    }    return path;}public List<Node> findPath(int sx, int sy, int dx, int dy) {    Node startNode = new Node(sx, sy);    Node endNode = new Node(dx, dy);    startNode.costG = 0;    startNode.costH = startNode.getManhattanCost(endNode);    startNode.parentNode = null;    openList.add(startNode);    while (!openList.isEmpty()) {        Node currentNode = (Node) openList.removeFirst();        if (currentNode == endNode) {            Log.d("android", "found path");            return buildPath(endNode);        } else {            currentNode.createNeighbors();            List<Node> neighbors = currentNode.getNeighbors();            for (int i = 0; i < neighbors.size(); i++) {                Node neighborNode = neighbors.get(i);                neighborNode.costG += currentNode.costG;                neighborNode.costH = neighborNode.getManhattanCost(endNode);                neighborNode.costF = neighborNode.costG + neighborNode.costH;                boolean isInOpen = openList.contains(neighborNode);                boolean isInClosed = closedList.contains(neighborNode);                if ((!isInOpen && !isInClosed) || neighborNode.costG < currentNode.costG) {                    neighborNode.parentNode = currentNode;                    neighborNode.costG += currentNode.costG;                    neighborNode.costF = neighborNode.costG + neighborNode.costH;                    if (isInClosed) {                        closedList.remove(neighborNode);                    }                    if (!isInOpen) {                        openList.add(neighborNode);                    }                }            }            closedList.add(currentNode);        }    }    openList.clear();    closedList.clear();    return null;}

}

public class Node {public Node parentNode;public List<Node> neighbors;public int x;public int y;public int costG;public int costH;public int costF;public Node(int x, int y) {    this.x = x;    this.y = y;    neighbors = new ArrayList<Node>();    costG = 10;}public void createNeighbors() {    neighbors.add(new Node(x + 1, y));    neighbors.add(new Node(x, y + 1));    neighbors.add(new Node(x - 1, y));    neighbors.add(new Node(x, y - 1));}public int getManhattanCost(Node node) {    int i = (int) Math.abs(x - node.x);    int j = (int) Math.abs(y - node.y);    costH = i + j;    return costH;}public int getTotalCost() {    return costG + costH;}public List<Node> getNeighbors() {    return neighbors;}

}

sx, sy, dx 和 dy 是二维数组中起始位置和目标位置。为了测试，我传入了固定的数字 sx = 1, sy = 1, dx = 5, dy = 5。换句话说，角色在 (1, 1) 处，触摸点在 (5, 5) 处。

回答：

经过大量努力后解决：在设置 currentNode 之后，清除 openList。