332. Reconstruct Itinerary
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题目
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.Note:
- 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.
Example 1:
Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]] Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]
Example 2:
Input: [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]] Output: ["JFK","ATL","JFK","SFO","ATL","SFO"] Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].But it is larger in lexical order.
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解题思路
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本题是关于图的边进行遍历,每张机票都是图的一条有向边,需要找出经过每条边的路径,并且必定有解本题,则对于某个节点(非起点)其只于一个节点相邻且只存在一条边,则这个节点必定是最后访问的,否则不可能遍历完所有边,并且这种点最多一个(不包含起点)。
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解法 1 –
DFS + 递归
- 解决步骤
- 将图建立起来,建立邻接表,使用
map<string, multiset<string>
来存储邻接表。使用multiset可以自动排序。(set的默认排序由小到大,multiset默认排序是由大到小
) - 从节点
JKF
开始DFS遍历,只要当前的映射集合multiset
里面还有节点,则取出这个节点,递归遍历这个节点,同时需要将这个节点从multiset
中删除掉,当映射集合multiset
为空的时候,则将节点加入到结果中 - 因为当前存储结果是回溯得到的,需要将结果的存储顺序反转输出
- 将图建立起来,建立邻接表,使用
- 实现代码
class Solution { public: vector<string> findItinerary(vector<pair<string, string>> tickets) { vector<string> v; map<string, multiset<string> > myMap; for(auto it : tickets) myMap[it.first].insert(it.second); dfs("JFK", v, myMap); reverse(v.begin(), v.end()); return v; } void dfs(string start, vector<string>& v, map<string, multiset<string> > &myMap) { while(myMap[start].size() > 0) { string next = *myMap[start].begin(); myMap[start].erase(myMap[start].begin()); dfs(next, v, myMap); } v.push_back(start); } };
- 解决步骤
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解法 2 –
DFS + 迭代
- 思路与解法一相同,利用数据结构
stack
进行迭代。 - 实现代码
class Solution { public: vector<string> findItinerary(vector<pair<string, string>> tickets) { vector<string> v; map<string, multiset<string> > myMap; for(auto it : tickets) myMap[it.first].insert(it.second); stack<string> myStack; myStack.push("JFK"); while(!myStack.empty()) { string node = myStack.top(); if(!myMap[node].size()) { myStack.pop(); v.push_back(node); } else { myStack.push(*myMap[node].begin()); myMap[node].erase(myMap[node].begin()); } } reverse(v.begin(), v.end()); return v; } };
- 思路与解法一相同,利用数据结构
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