[LeetCode] 332. Reconstruct Itinerary

题目描述

Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK. Thus, the itinerary must begin with JFK.

注意

• If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string.
• For example, the itinerary [“JFK”, “LGA”] has a smaller lexical order than [“JFK”, “LGB”].
• All airports are represented by three capital letters (IATA code).
• You may assume all tickets form at least one valid itinerary.

分析

开始的想法就是在数组中寻找符合from的to中最小的加入数组中，但是发现有个测例是不通过的，在该测例中这样找无法找到后续的to。

class Solution {public:  vector<string> findItinerary(vector<pair<string, string>> tickets) {    const int n = tickets.size();    vector<string> result;    result.push_back("JFK");    int visited[n];    int i, j, pos, size = tickets.size(), count = 0;    string from = "JFK", to;    for (i = 0; i < size; i++) {      visited[i] = 0;    }    while (count != size) {      to = "";      for (j = 0; j < size; j++) {        if (visited[j]) continue;        if (get<0>(tickets[j]) == from) {          if (to == "") {            to = get<1>(tickets[j]);            pos = j;          } else {            if (to > get<1>(tickets[j])) {              to = get<1>(tickets[j]);              pos = j;            }          }        }      }      count++;      visited[pos] = 1;      from = to;      result.push_back(to);    }    return result;  }};

后来从上网查找思路，找到一个思路是建立一个以from为key的哈希表，然后按照字典序进行DFS查找，相同字典序则从小到大进行选择，直到某处没有可以相连的to那么就找到了想要的路径，这样才把结果依次加入数组，这样数组的信息是反向的，之后对这个路径进行反向输出就可以了。

代码

class Solution {public:  void DFS(string from) {    while (hashTable[from].size() > 0) {      string to = *hashTable[from].begin();      hashTable[from].erase(hashTable[from].begin());      DFS(to);    }    result.push_back(from);  }  vector<string> findItinerary(vector<pair<string, string>> tickets) {    if(tickets.size() == 0) return {};    for (auto val : tickets) {      hashTable[val.first].insert(val.second);    }    DFS("JFK");    reverse(result.begin(), result.end());    return result;  }private:  vector<string> result;  unordered_map<string, multiset<string>> hashTable;};