paths.remove(paths.size() - 1);
时间: 2024-04-10 19:28:17 浏览: 20
根据你的代码片段 `paths.remove(paths.size() - 1)`,它似乎是在一个具有索引的数据结构(例如列表、数组)中删除最后一个元素。
然而,Python 中的列表没有 `size()` 方法,而是使用 `len()` 函数来获取列表的长度。因此,你可以使用 `len(paths)` 来获取列表 `paths` 的长度。
如果你想要删除列表中的最后一个元素,可以使用以下代码:
```python
paths = ['path1', 'path2', 'path3']
paths.remove(paths[len(paths) - 1])
print(paths) # 输出: ['path1', 'path2']
```
在这个示例中,我们通过调用 `remove` 方法来删除列表 `paths` 中的最后一个元素。我们使用 `len(paths) - 1` 来获取最后一个元素的索引,并将其作为参数传递给 `remove` 方法。最后,我们打印出更新后的列表 `paths`,结果为 `['path1', 'path2']`。
相关问题
while any(openSet(:) > 0) % Find the minimum fScore within the open set [~, current] = min(fScore(:)); % If we've reached the goal if current == goal % Get the full path and return it final = get_path(cameFrom, current); return end % Linear index -> row, col subscripts rc = rem(current - 1, mapSize(1)) + 1; cc = (current - rc) / mapSize(1) + 1; % Remove CURRENT from openSet openSet(rc, cc) = false; % Place CURRENT in closedSet closedSet(rc, cc) = true; fScore(rc, cc) = inf; gScoreCurrent = gScore(rc, cc) + costs(rc, cc); % Get all neighbors of CURRENT. Neighbors are adjacent indices on % the map, including diagonals. % Col 1 = Row, Col 2 = Col, Col 3 = Distance to the neighbor n_ss = [ ... rc + 1, cc + 1, S2 ; ... rc + 1, cc + 0, 1 ; ... rc + 1, cc - 1, S2 ; ... rc + 0, cc - 1, 1 ; ... rc - 1, cc - 1, S2 ; ... rc - 1, cc - 0, 1 ; ... rc - 1, cc + 1, S2 ; ... rc - 0, cc + 1, 1 ; ... ]; % keep valid indices only valid_row = n_ss(:,1) >= 1 & n_ss(:,1) <= mapSize(1); valid_col = n_ss(:,2) >= 1 & n_ss(:,2) <= mapSize(2); n_ss = n_ss(valid_row & valid_col, :); % subscripts -> linear indices neighbors = n_ss(:,1) + (n_ss(:,2) - 1) .* mapSize(1); % only keep neighbors in the map and not in the closed set ixInMap = map(neighbors) & ~closedSet(neighbors); neighbors = neighbors(ixInMap); % distance to each kept neighbor dists = n_ss(ixInMap, 3); % Add each neighbor to the open set openSet(neighbors) = true; % TENTATIVE_GSCORE is the score from START to NEIGHBOR. tentative_gscores = gScoreCurrent + costs(neighbors) .* dists; % IXBETTER indicates where a better path was found ixBetter = tentative_gscores < gScore(neighbors); bestNeighbors = neighbors(ixBetter); % For the better paths, update scores cameFrom(bestNeighbors) = current; gScore(bestNeighbors) = tentative_gscores(ixBetter); fScore(bestNeighbors) = gScore(bestNeighbors) + compute_cost(mapSize, bestNeighbors, gr, gc); end % while end
这段代码是 A* 算法的核心实现部分,用于寻找两点之间的最短路径。具体来说,它依次对起点周围的格子进行探索,计算每个格子到起点的代价(gScore)以及到终点的估计代价(fScore),并将其加入 openSet 中。然后从 openSet 中选取 fScore 最小的格子进行探索,直到找到终点或 openSet 中没有格子可供探索为止。
在探索每个格子时,首先将其从 openSet 中移除并加入 closedSet 中,然后计算该格子与周围格子的代价,并将未被探索过的格子加入 openSet 中。如果发现了新的更优路径,就更新该格子到起点的代价和 fScore,并将其加入到 cameFrom 列表中,表示它是从哪个格子转移而来的。
最终,如果找到了终点,就从 cameFrom 列表中回溯路径。
二叉树BinaryTree.java类实现以下方法:输出所有直径及其路径长度 public static<T> void diameterAll(BinaryTree<T> bitree)
以下是一个简单的二叉树实现的Java代码,包含了输出所有直径及其路径长度的方法:
```java
public class BinaryTree<T> {
private Node<T> root;
// 构造函数
public BinaryTree(Node<T> root) {
this.root = root;
}
// 节点类
private static class Node<T> {
private T data;
private Node<T> left;
private Node<T> right;
public Node(T data) {
this.data = data;
this.left = null;
this.right = null;
}
}
// 输出所有直径及其路径长度
public static <T> void diameterAll(BinaryTree<T> bitree) {
if (bitree.root == null) {
System.out.println("Binary tree is empty.");
return;
}
List<List<Node<T>>> paths = new ArrayList<>();
List<Integer> diameters = new ArrayList<>();
findPaths(bitree.root, paths, new ArrayList<>());
calculateDiameters(bitree.root, paths, diameters);
for (int i = 0; i < diameters.size(); i++) {
System.out.println("Diameter: " + diameters.get(i) + ", Path: ");
for (Node<T> node : paths.get(i)) {
System.out.print(node.data + " ");
}
System.out.println();
}
}
// 查找所有路径
private static <T> void findPaths(Node<T> node, List<List<Node<T>>> paths, List<Node<T>> path) {
if (node == null) {
return;
}
path.add(node);
if (node.left == null && node.right == null) {
paths.add(new ArrayList<>(path));
} else {
findPaths(node.left, paths, path);
findPaths(node.right, paths, path);
}
path.remove(path.size() - 1);
}
// 计算直径
private static <T> int calculateDiameters(Node<T> node, List<List<Node<T>>> paths, List<Integer> diameters) {
if (node == null) {
return 0;
}
int leftHeight = calculateDiameters(node.left, paths, diameters);
int rightHeight = calculateDiameters(node.right, paths, diameters);
int diameter = leftHeight + rightHeight;
diameters.add(diameter);
return Math.max(leftHeight, rightHeight) + 1;
}
public static void main(String[] args) {
// 创建二叉树示例
Node<Integer> node1 = new Node<>(1);
Node<Integer> node2 = new Node<>(2);
Node<Integer> node3 = new Node<>(3);
Node<Integer> node4 = new Node<>(4);
Node<Integer> node5 = new Node<>(5);
node1.left = node2;
node1.right = node3;
node2.left = node4;
node3.right = node5;
BinaryTree<Integer> bitree = new BinaryTree<>(node1);
// 输出所有直径及其路径长度
diameterAll(bitree);
}
}
```
这段代码使用了二叉树的先序遍历来查找所有路径,然后计算每个路径的直径。最后,输出每个直径及其路径长度。以上是一个简单的实现,你可以根据自己的需求进行修改和扩展。