import heapq import copy # 定义状态类 class State: def __init__(self, board, moves=0, parent=None, last_move=None): self.board = board self.moves = moves self.parent = parent self.last_move = last_move def __lt__(self, other): return self.moves < other.moves def __eq__(self, other): return self.board == other.board # 定义转移函数 def move(state, direction): new_board = copy.deepcopy(state.board) for i in range(len(new_board)): if 0 in new_board[i]: j = new_board[i].index(0) break if direction == "up": if i == 0: return None else: new_board[i][j], new_board[i-1][j] = new_board[i-1][j], new_board[i][j] elif direction == "down": if i == len(new_board)-1: return None else: new_board[i][j], new_board[i+1][j] = new_board[i+1][j], new_board[i][j] elif direction == "left": if j == 0: return None else: new_board[i][j], new_board[i][j-1] = new_board[i][j-1], new_board[i][j] elif direction == "right": if j == len(new_board)-1: return None else: new_board[i][j], new_board[i][j+1] = new_board[i][j+1], new_board[i][j] return State(new_board, state.moves+1, state, direction) # 定义A*算法 def astar(start, goal): heap = [] closed = set() heapq.heappush(heap, start) while heap: state = heapq.heappop(heap) if state.board == goal: path = [] while state.parent: path.append(state) state = state.parent path.append(state) return path[::-1] closed.add(state) for direction in ["up", "down", "left", "right"]: child = move(state, direction) if child is None: continue if child in closed: continue if child not in heap: heapq.heappush(heap, child) else: for i, (p, c) in enumerate(heap): if c == child and p.moves > child.moves: heap[i] = (child, child) heapq.heapify(heap) # 测试 start_board = [[1, 2, 3], [4, 5, 6], [7, 8, 0]] goal_board = [[2, 3, 6], [1, 5, 8], [4, 7, 0]] start_state = State(start_board) goal_state = State(goal_board) path = astar(start_state, goal_board) for state in path: print(state.board)

ASTAR.rar_Astar算法_a star_path planning_python 路径

在Python中，A*算法通常通过一个优先队列（如heapq模块中的heappush和heappop）来管理待处理节点，根据F(n)值进行排序。同时，还需要一个哈希表来存储已经访问过的节点，避免重复探索。在"ASTAR.py"文件中，...

python-source-datastructure.pdf.zip_Work It

5. **堆队列(HeapQueue)**：Python的heapq模块实现了最小堆数据结构，常用于优先队列。堆是一种特殊的树形数据结构，满足堆性质：父节点的值总是小于或等于（最大堆）或大于或等于（最小堆）其子节点的值。 6. **...

class FixedSizeHeap: def init(self, max_size): self.max_size = max_size self.heap = [] def push(self, item): if len(self.heap) < self.max_size: heapq.heappush(self.heap, item) else: min_item = heapq.heappop(self.heap) if item > min_item: heapq.heappush(self.heap, item) else: heapq.heappush(self.heap, min_item) def pop(self): return heapq.heappop(self.heap) def len(self): return len(self.heap)对于这样的类怎么打印类中的元素

def __init__(self, max_size): self.max_size = max_size self.heap = [] def push(self, item): if len(self.heap) < self.max_size: heapq.heappush(self.heap, item) else: min_item = heapq.heappop...

class KthLargest: def init(self, k: int, nums: List[int]): self.k = k hp = [] for x in nums: heapq.heappush(hp,x) if len(hp) > k: heappop(hp) self.hp = hp def add(self, val: int) -> int: heappush(self.hp,val) if len(self.hp)>self.k: heappop(self.hp) return self.hp[0]

这是一个实现 Kth Largest Element in a Stream 的类，使用了 Python 的 heapq 库（堆）。在类的初始化方法 __init__ 中，首先将 nums 中的元素依次添加到一个最小堆 hp 中，如果 hp 中的元素数量超过了 k，则弹出...

class Path(object): def init(self,path,distancecost,timecost): self.path = path self.distancecost = distancecost self.timecost = timecost #路径上最后一个节点 def getLastNode(self): return self.path[-1] #获取路径路径 @property def path(self): return self.path #判断node是否为路径上最后一个节点 def isLastNode(self, node): return node == self.getLastNode() #增加加点和成本产生一个新的path对象 def addNode(self, node, dprice, tprice): return Path(self.path+[node],self.distancecost + dprice,self.timecost + tprice) #输出当前路径 def printPath(self): for n in self.path: if self.isLastNode(node=n): print(n) else: print(n, end="->") print(f"最短路径距离(self.distancecost:.0f)m") print(f"红绿路灯个数（self.timecost:.0f）个") #获取路径总成本的只读属性 @property def dCost(self): return self.distancecost @property def tCost(self): return self.timecost class DirectedGraph(object): def init(self, d): if isinstance(d, dict): self.graph = d else: self.graph = dict() print('Sth error') #通过递归生成所有可能的路径 def generatePath(self, graph, path, end, results, distancecostIndex, timecostIndex): current = path.getLastNode() if current == end: results.append(path) else: for n in graph[current]: if n not in path.path: self.generatePath(graph, path.addNode(n,self.graph[path.getLastNode()][n][distancecostIndex][timecostIndex]), end, results, distancecostIndex, timecostIndex) #搜索start到end之间时间或空间最短的路径，并输出 def searchPath(self, start, end, distancecostIndex, timecostIndex): results = [] self.generatePath(self.graph, Path([start],0,0), end, results,distancecostIndex,timecostIndex) results.sort(key=lambda p: p.distanceCost) results.sort(key=lambda p: p.timeCost) print('The {} shortest path from '.format("spatially" if distancecostIndex==0 else "temporally"), start, ' to ', end, ' is:', end="") print('The {} shortest path from '.format("spatially" if timecostIndex==0 else "temporally"), start, ' to ', end, ' is:', end="") results[0].printPath() #调用searchPath搜索start到end之间的空间最短的路径，并输出 def searchSpatialMinPath(self,start, end): self.searchPath(start,end,0,0) #调用searc 优化这个代码

def __init__(self, path = [], distance_cost = 0, time_cost = 0): self._path = path self._distance_cost = distance_cost self._time_cost = time_cost def get_last_node(self): return self._path[-1...

import heapqclass Graph: def init(self, vertices): self.vertices = vertices self.adjacency_list = {v: [] for v in vertices} def add_edge(self, u, v, weight): self.adjacency_list[u].append((v, weight)) self.adjacency_list[v].append((u, weight)) def dijkstra(self, source): heap = [(0, source)] visited = set() shortest_distances = {v: float('inf') for v in self.vertices} shortest_distances[source] = 0 while heap: (dist, current_vertex) = heapq.heappop(heap) if current_vertex in visited: continue visited.add(current_vertex) for (neighbor, weight) in self.adjacency_list[current_vertex]: distance = dist + weight if distance < shortest_distances[neighbor]: shortest_distances[neighbor] = distance heapq.heappush(heap, (distance, neighbor)) for v in self.vertices: print(f"{source}->{v}的最短路径为：{v} 长度：{shortest_distances[v]}")# 测试代码vertices = ['A', 'B', 'C', 'D', 'E']graph = Graph(vertices)graph.add_edge('A', 'B', 6)graph.add_edge('A', 'D', 1)graph.add_edge('B', 'C', 5)graph.add_edge('B', 'D', 2)graph.add_edge('B', 'E', 2)graph.add_edge('C', 'E', 5)graph.add_edge('D', 'E', 1)graph.dijkstra('A')

这是一个使用 Dijkstra 算法求解最短路径的代码。假设有一个无向图，使用字典 adjacency_list 存储每个顶点的邻接表。在 dijkstra 方法中，首先将起始顶点放入堆中，然后使用 visited 集合记录已经访问过的...

class randomMAP: def init(self,size,start,goal,p): self.size=size self.start=start self.goal=goal self.p=p def creatMAP(self): self.map=np.zeros(self.size,dtype='int') for i in range(self.map.shape[0]): for j in range(self.map.shape[1]): if((i!=self.start[0] or i!=self.start[1]) and (j!=self.goal[0] or j!=self.goal[1])) and random.random() <self.p: self.map[i][j] = 5 map1=randomMAP((20,20),(0,0),(19,19),0.3) map1.creatMAP() plt.matshow(map1.map) plt.show()在这个随机地图的基础上创建Point和Astar类，用于A*算法解决迷宫问题

while len(open_list) > 0: current_point = heapq.heappop(open_list) if current_point.x == goal_point.x and current_point.y == goal_point.y: path = [] while current_point.parent is not None: path....

为这段代码添加中文释义import heapq from collections import defaultdict def huffman_encoding(data): # 计算字符频率 freq = defaultdict(int) for char in data: freq[char] += 1 # 将频率转化为堆 heap = [[weight, [char, ""]] for char, weight in freq.items()] heapq.heapify(heap) # 合并堆中的节点，生成霍夫曼编码树 while len(heap) > 1: low = heapq.heappop(heap) high = heapq.heappop(heap) for pair in low[1:]: pair[1] = '0' + pair[1] for pair in high[1:]: pair[1] = '1' + pair[1] heapq.heappush(heap, [low[0] + high[0]] + low[1:] + high[1:]) # 生成霍夫曼编码表 code_table = dict(heapq.heappop(heap)[1:]) # 编码 encoded_data = "" for char in data: encoded_data += code_table[char] return encoded_data, code_table def huffman_decoding(encoded_data, code_table): # 将编码表反转，方便解码 reverse_code_table = {v: k for k, v in code_table.items()} # 解码 decoded_data = "" code = "" for bit in encoded_data: code += bit if code in reverse_code_table: decoded_data += reverse_code_table[code] code = "" return decoded_data data = "hello world" encoded_data, code_table = huffman_encoding(data) print("Encoded data:", encoded_data) print("Code table:", code_table) decoded_data = huffman_decoding(encoded_data, code_table) print("Decoded data:", decoded_data)

import heapq from collections import defaultdict def huffman_encoding(data): # 计算字符频率 freq = defaultdict(int) # 使用 defaultdict 来实现字符频率的计算 for char in data: freq[char] += 1 # ...

优化这段代码 import heapq import numpy as np def MinTimeSlot(containerList, req, CONTAINER_SPARE): seq1 = 0 spareContainerList = [] for container in containerList: if container.appId == req.appId and container.state == CONTAINER_SPARE: seq1 += 1 heapq.heappush(spareContainerList, (-1 * container.killedTime, seq1, container)) return spareContainerList

def __init__(self, containerList): self.containerDict = {} for container in containerList: if container.appId not in self.containerDict: self.containerDict[container.appId] = [] if container....

import random import heapq # 生成无向图 def generate_graph(n, p): graph = [[0] * n for _ in range(n)] for i in range(n): for j in range(i+1, n): if random.random() < p: graph[i][j] = graph[j][i] = random.randint(1, 10) return graph # Prim算法求最小生成树 def prim(graph): n = len(graph) visited = [False] * n heap = [(0, 0)] mst = [] while heap: weight, node = heapq.heappop(heap) if visited[node]: continue visited[node] = True mst.append((weight, node)) for i in range(n): if not visited[i] and graph[node][i] > 0: heapq.heappush(heap, (graph[node][i], i)) return mst # Kruskal算法求最小生成树 def kruskal(graph): n = len(graph) edges = [] for i in range(n): for j in range(i+1, n): if graph[i][j] > 0: edges.append((graph[i][j], i, j)) edges.sort() parent = list(range(n)) mst = [] for weight, u, v in edges: pu, pv = find(parent, u), find(parent, v) if pu != pv: mst.append((weight, u, v)) parent[pu] = pv return mst def find(parent, x): if parent[x] != x: parent[x] = find(parent, parent[x]) return parent[x] # 生成图 graph = generate_graph(10, 0.6) print(graph) mst_prim = prim(graph) print("Prim算法求最小生成树：", mst_prim) mst_kruskal = kruskal(graph) print("Kruskal算法求最小生成树：", mst_kruskal) # Dijkstra算法求最短路径 def dijkstra(graph, start, end): n = len(graph) dist = [float('inf')] * n dist[start] = 0 visited = [False] * n heap = [(0, start)] while heap: d, u = heapq.heappop(heap) if visited[u]: continue visited[u] = True for v in range(n): if graph[u][v] > 0: if dist[u] + graph[u][v] < dist[v]: dist[v] = dist[u] + graph[u][v] heapq.heappush(heap, (dist[v], v)) return dist[end] # Bellman-Ford算法求最短路代码分析

这段代码主要实现了生成无向图、Prim算法和Kruskal算法求最小生成树、Dijkstra算法和Bellman-Ford算法求最短路径。具体分析如下： 1. 生成无向图：通过调用generate_graph函数，可以生成一个无向图，其中n表示节点...

import Astar import heapq start_cor = (19, 0) waypoints = [(5, 15), (5, 1), (9, 3), (11, 17), (7, 19), (15, 19), (13, 1), (15, 5)] end_cor = (1, 20) def distance(_from, _to): x1, y1 = _from x2, y2 = _to distancepath = Astar.find_path(x1, y1, x2, y2) return distancepath n = len(waypoints) adj_matrix = [[0] * n for _ in range(n)] for i in range(n): for j in range(i + 1, n): dist = distance(waypoints[i], waypoints[j]) adj_matrix[i][j] = dist adj_matrix[j][i] = dist start = 0 end = n - 1 distances = [[float('inf')] * (n + 1) for _ in range(n)] visited = set() heap = [(0, 0, start)] while heap: (dist, num_visited, current) = heapq.heappop(heap) if current == end and num_visited == 8: break if (current, num_visited) in visited: continue visited.add((current, num_visited)) for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0: new_num_visited = num_visited if neighbor in range(start + 1, end) and (current not in range(start + 1, end)) and num_visited < 8: new_num_visited += 1 new_distance = dist + weight if new_distance < distances[neighbor][new_num_visited]: distances[neighbor][new_num_visited] = new_distance heapq.heappush(heap, (new_distance, new_num_visited, neighbor)) min_dist = float('inf') min_num_visited = 8 for i in range(8): if distances[end][i] < min_dist: min_dist = distances[end][i] min_num_visited = i path = [end] current = end num_visited = min_num_visited for i in range(len(waypoints), 0, -1): if current in range(i): num_visited -= 1 for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0 and (neighbor, num_visited) in visited and distances[neighbor][num_visited] + weight == \ distances[current][num_visited]: path.append(neighbor) current = neighbor break path.reverse() print(f"The optimal path from start to end through the 8 waypoints is: {path}") print(f"The total distance is: {distances[end][min_num_visited]}")

这段代码是用来解决一个路径规划问题，目标是找到从起点到终点经过8个给定点的最短路径。其中使用了A*算法来计算两点之间的距离，并利用堆来实现最小优先队列。具体流程是先计算出8个给定点之间的距离，再从起点开始...

import heapq def min_refuel_stops(n: int, stations: list[int], L: int) -> int: stops = 0 # 记录加油次数 last_pos = 0 # 上一个加油站的位置 tank = L # 当前油量 pq = [] # 定义一个小根堆，用于选择最大的油量 while True: # 当前油可以到达终点，直接返回加油次数 if tank >= n - last_pos: return stops # 遍历所有加油站，找到当前能够到达的加油站中油量最多的那一个 while stations and stations[0] <= last_pos + tank: heapq.heappush(pq, -stations.pop(0)) # 如果当前无法到达任何一个加油站，则无法到达终点，返回-1 if not pq: return -1 # 在当前能够到达的加油站中选择油量最大的那一个 tank += -heapq.heappop(pq) stops += 1 # 更新上一个加油站的位置 last_pos = stations[0] if stations else n return -1 n = 500 stations = [50, 100, 150, 200, 250, 300, 350, 400, 450, 500, 550, 600, 650, 700, 750, 800, 850, 900, 950] L = 100 print(min_refuel_stops(n, stations, L)) # 2 修改上述代码并输出

import heapq def min_refuel_stops(n: int, stations: list[int], L: int) -> int: stops = 0 # 记录加油次数 last_pos = 0 # 上一个加油站的位置 tank = L # 当前油量 pq = [] # 定义一个小根堆，用于选择...

start_cor = (19, 0)waypoints = [(5, 15), (5, 1), (9, 3), (11, 17), (7, 19), (15, 19), (13, 1), (15, 5)] end_cor = (1, 20) def distance(_from, _to): x1, y1 = _from x2, y2 = _to distancepath = Astar.find_path(x1, y1, x2, y2) return distancepath n = len(waypoints) adj_matrix = [[0] * n for _ in range(n)] for i in range(n): for j in range(i + 1, n): dist = distance(waypoints[i], waypoints[j]) adj_matrix[i][j] = dist adj_matrix[j][i] = dist start = 0 end = n - 1 distances = [[float('inf')] * (n + 1) for _ in range(n)] visited = set() heap = [(0, 0, start)] while heap: (dist, num_visited, current) = heapq.heappop(heap) if current == end and num_visited == 8: break if (current, num_visited) in visited: continue visited.add((current, num_visited)) for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0: new_num_visited = num_visited if neighbor in range(start + 1, end) and (current not in range(start + 1, end)) and num_visited < 8: new_num_visited += 1 new_distance = dist + weight if new_distance < distances[neighbor][new_num_visited]: distances[neighbor][new_num_visited] = new_distance heapq.heappush(heap, (new_distance, new_num_visited, neighbor)) min_dist = float('inf') min_num_visited = 8 for i in range(8): if distances[end][i] < min_dist: min_dist = distances[end][i] min_num_visited = i每排是什么意思帮我加上注释

if weight > 0: new_num_visited = num_visited # 如果邻居节点是路标点，并且当前节点不是路标点，并且途经的路标点个数还没有达到 8 个，则途经的路标点个数加 1 if neighbor in range(start + 1, end) and ...

waypoints = [(5, 15), (5, 1), (9, 3), (11, 17), (7, 19), (15, 19), (13, 1), (15, 5)] end_cor = (1, 20) def distance(_from, _to): x1, y1 = _from x2, y2 = _to distancepath = Astar.find_path(x1, y1, x2, y2) return distancepath n = len(waypoints) adj_matrix = [[0] * n for _ in range(n)] for i in range(n): for j in range(i + 1, n): dist = distance(waypoints[i], waypoints[j]) adj_matrix[i][j] = dist adj_matrix[j][i] = dist start = 0 end = n - 1 distances = [[float('inf')] * (n + 1) for _ in range(n)] visited = set() heap = [(0, 0, start)] while heap: (dist, num_visited, current) = heapq.heappop(heap) if current == end and num_visited == 8: break if (current, num_visited) in visited: continue visited.add((current, num_visited)) for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0: new_num_visited = num_visited if neighbor in range(start + 1, end) and (current not in range(start + 1, end)) and num_visited < 8: new_num_visited += 1 new_distance = dist + weight if new_distance < distances[neighbor][new_num_visited]: distances[neighbor][new_num_visited] = new_distance heapq.heappush(heap, (new_distance, new_num_visited, neighbor)) min_dist = float('inf') min_num_visited = 8 for i in range(8): if distances[end][i] < min_dist: min_dist = distances[end][i] min_num_visited = i每排是什么意思

这段代码实现了一个寻找最短路径的算法，其中： - waypoints 是一个包含多个点坐标的列表，代表路径上的多个路标点。 - end_cor 是路径的终点坐标。 - distance 函数用于计算两个点之间的距离，使用了 Astar 算法...

import Astar import heapq start_cor = (19, 0) treasures = [(5, 15), (5, 1), (9, 3), (11, 17), (7, 19), (15, 19), (13, 1), (15, 5)] end_cor = (1, 20) # 定义一个函数计算两个坐标之间的距离 def distance(_from, _to): # 返回从起点到终点的最短路径 x1, y1 = _from x2, y2 = _to distancepath = Astar.find_path(x1, y1, x2, y2) return distancepath n = len(treasures) adj_matrix = [[0] * n for _ in range(n)] for i in range(n): for j in range(i + 1, n): dist = distance(treasures[i], treasures[j]) adj_matrix[i][j] = dist adj_matrix[j][i] = dist # 使用Dijkstra算法求解最短路径 start = 0 end = n - 1 distances = [float('inf')] * n distances[start] = 0 visited = set() heap = [(0, start)] while heap: (dist, current) = heapq.heappop(heap) if current == end: break if current in visited: continue visited.add(current) for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0 and neighbor not in visited: new_distance = dist + weight if new_distance < distances[neighbor]: distances[neighbor] = new_distance heapq.heappush(heap, (new_distance, neighbor)) # 输出结果 path = [end] current = end while current != start: for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0 and distances[current] == distances[neighbor] + weight: path.append(neighbor) current = neighbor break print(path) path.reverse() print(f"从第{start+1}个坐标开始经过其他几个坐标最后到达第{end+1}个坐标的最短路线为：{path}") print(f"总距离为：{distances[end]}"

1. 导入 Astar 模块和 heapq 模块。 2. 定义起点坐标 start_cor 和终点坐标 end_cor，以及一个包含多个宝藏坐标的列表 treasures。 3. 定义一个函数 distance，用于计算两个坐标之间的距离。在该函数中，...

import Astar import heapq start = (19, 0) treasures = [(5, 15), (5, 1), (9, 3), (11, 17), (7, 19), (15, 19), (13, 1), (15, 5)] end = (1, 20) # 定义一个函数计算两个坐标之间的距离 def distance(_from, _to): # 返回从起点到终点的最短路径 x1, y1 = _from x2, y2 = _to distancepath = Astar.find_path(x1, y1, x2, y2) return distancepath n = len(treasures) adj_matrix = [[0] * n for _ in range(n)] for i in range(n): for j in range(i + 1, n): dist = distance(treasures[i], treasures[j]) adj_matrix[i][j] = dist adj_matrix[j][i] = dist # 使用Dijkstra算法求解最短路径 start = 0 end = n - 1 distances = [float('inf')] * n distances[start] = 0 visited = set() heap = [(0, start)] while heap: (dist, current) = heapq.heappop(heap) if current == end: break if current in visited: continue visited.add(current) for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0 and neighbor not in visited: new_distance = dist + weight if new_distance < distances[neighbor]: distances[neighbor] = new_distance heapq.heappush(heap, (new_distance, neighbor)) # 输出结果 path = [end] current = end while current != start: for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0 and distances[current] == distances[neighbor] + weight: path.append(neighbor) current = neighbor break print(path) path.reverse() print(f"从第{start+1}个坐标开始经过其他几个坐标最后到达第{end+1}个坐标的最短路线为：{path}") print(f"总距离为：{distances[end]}")

1. 导入 Astar 模块和 heapq 模块。 2. 定义起点 start 和终点 end 的坐标，以及一个包含多个宝藏坐标的列表 treasures。 3. 定义一个函数 distance，用于计算两个坐标之间的距离。在该函数中，调用了 A* ...

import heapqdef prim(graph, start): n = len(graph) visited = [False] * n distance = [float('inf')] * n distance[start] = 0 heap = [(0, start)] while heap: _, u = heapq.heappop(heap) if visited[u]: continue visited[u] = True for v, w in graph[u]: if not visited[v] and w < distance[v]: distance[v] = w heapq.heappush(heap, (w, v)) return sum(distance)def min_distance_sum(graph): n = len(graph) min_distance = float('inf') for i in range(n): distance_sum = prim(graph, i) min_distance = min(min_distance, distance_sum) return min_distancegraph = [ [(1, 2), (2, 1)], [(0, 2), (2, 3), (3, 1)], [(0, 1), (1, 3), (3, 1)], [(1, 1), (2, 1)]]print(min_distance_sum(graph)) # 输出：4把这段代码转换成c++

distance[start] = 0; priority_queue, vector, greater<pii>> pq; pq.push({0, start}); while (!pq.empty()) { int u = pq.top().second; pq.pop(); if (visited[u]) { continue; } visited[u] = true; ...

File "D:\23101\比赛\光电赛\maze_car\测试\11111.py", line 46, in dijkstra heapq.heappush(heap, (new_distance, neighbor)) TypeError: '<' not supported between instances of 'Node' and 'Node'错误如上代码如下 def dijkstra(self, start_node): start_node.distance = 0 heap = [(start_node.distance, start_node)] while heap: (distance, node) = heapq.heappop(heap) if node.visited: continue node.visited = True for (neighbor, edge_distance) in node.neighbors: if not neighbor.visited: new_distance = node.distance + edge_distance if new_distance < neighbor.distance: neighbor.distance = new_distance heapq.heappush(heap, (new_distance, neighbor))

这个错误是由于在使用 heapq 堆操作时，Python 不知道如何比较节点（Node）对象。您需要在 Node 类中实现 __lt__ 方法，以便 Python 可以比较这些对象。例如： python class Node: def __init__(self...

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class KthLargest: def init(self, k: int, nums: List[int]): self.k = k hp = [] for x in nums: heapq.heappush(hp,x) if len(hp) > k: heappop(hp) self.hp = hp def add(self, val: int) -> int: heappush(self.hp,val) if len(self.hp)>self.k: heappop(self.hp) return self.hp[0]