Have a HashMap<String, TreeMap<Integer, String>> hm to store keys. The value is TreeMap sorted based on timestamp. set, update hm. get, first get the TreeMap based on key, then use floorKey to find the largest key <= timestamp. Time Complexity: set, O(logn). get, O(logn). n...
Time Complexity: O(E+VlogV). It takes O(E) time to construct graph. O(VlogV) time to traverse all the points. Space: O(E+V). O(E) for graph, O(V) for minHeap and Set. AC Java: 1classSolution {2publicintnetworkDelayTime(int[][] times,intN,intK) {3Map<Integer, List<int...
Background and motivation This data container is an analogue of LinkedHashMap implementation in Java, which is missing in C#. In the latest release was finally added PriorityQueue which was in Java since 2004. Bridging the gap between AP...
思路: Time Complexity: O() Space Complexity: O() Solution Code: classSolution{public intnetworkDelayTime(int[][]times,int N,int K){if(times==null||times.length==0){return-1;}// graph initMap<Integer,Map<Integer,Integer>>graph=newHashMap<>();for(int[]time:times){if(!graph.contain...
但我不确定是否应该考虑在使用 java.util 实现的大型 hashmap 中可能出现的冲突问题。 - a_confused_student 6 它的时间复杂度是O(1)摊销,就像containsKey、put或remove一样。如果不进行摊销分析,它可能是O(n),因为任何更改都可能触发重新哈希,潜在地涉及所有条目。但是没有人关心非摊销分析。 - Zabuzard 我想...
aliasMap.get(q); if (aliasMap == null) { aliasMap = Maps.newHashMap(); this.aliasMap.put(q, aliasMap); } String alias = aliased.getAlias(); if (alias == null) { alias = aliased.getQualified().getSegments().getLast(); from.alias(alias); } if (aliasMap.containsKey(alias))...
// Network Delay Time// Dijkstra// Time Complexity: O(ElogN), Space Complexity: O(N + E)classSolution{publicintnetworkDelayTime(int[][]times,intN,intK){// adjacency list, map<vertex_id, map<vertex_id, weight>>Map<Integer,Map<Integer,Integer>>graph=newHashMap<>();for(int[]time:ti...