[LeetCode] 332. Reconstruct Itinerary

最新推荐文章于 2019-07-06 22:00:19 发布

原创最新推荐文章于 2019-07-06 22:00:19 发布 · 276 阅读

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#DFS #leetcode

本文介绍了一种解决行程重建问题的方法，即根据一系列机票信息重构旅行路线。重点在于如何使用哈希表进行字典序DFS搜索来确保正确且高效地构建行程路径。

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[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;
};