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题目:(中等)
标签:图、Prim算法、Kruskal算法
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
---|---|---|---|
Ans 1 (Python) | O ( N l o g N ) O(NlogN) O(NlogN) | O ( N ) O(N) O(N) | 208ms (62.25%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一:
class Solution: def minimumCost(self, n: int, connections: List[List[int]]) -> int: min_idx, min_val = (-1, -1), float("inf") graph = collections.defaultdict(dict) for edge in connections: if edge[1] not in graph[edge[0]] or edge[2] < graph[edge[0]][edge[1]]: graph[edge[0]][edge[1]] = edge[2] graph[edge[1]][edge[0]] = edge[2] if edge[2] < min_val: min_idx, min_val = (edge[0], edge[1]), edge[2] ans = 0 waiting = { i for i in range(1, n + 1)} heap = [] # 添加最短的边 ans += min_val waiting.remove(min_idx[0]) waiting.remove(min_idx[1]) for n2, v2 in graph[min_idx[0]].items(): if n2 in waiting: heapq.heappush(heap, (v2, n2)) for n2, v2 in graph[min_idx[1]].items(): if n2 in waiting: heapq.heappush(heap, (v2, n2)) while heap and waiting: v1, n1 = heapq.heappop(heap) if n1 in waiting: ans += v1 waiting.remove(n1) for n2, v2 in graph[n1].items(): if n2 in waiting: heapq.heappush(heap, (v2, n2)) return ans if not waiting else -1
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