# [LeetCode] Reconstruct Itinerary 重建行程单

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:

1. 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"].
2. All airports are represented by three capital letters (IATA code).
3. You may assume all tickets may form at least one valid itinerary.

Example 1:
tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Return ["JFK", "MUC", "LHR", "SFO", "SJC"].

Example 2:
tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Return ["JFK","ATL","JFK","SFO","ATL","SFO"].
Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"]. But it is larger in lexical order.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

class Solution {
public:
vector<string> findItinerary(vector<pair<string, string>> tickets) {
vector<string> res;
unordered_map<string, multiset<string>> m;
for (auto a : tickets) {
m[a.first].insert(a.second);
}
dfs(m, "JFK", res);
return vector<string> (res.rbegin(), res.rend());
}
void dfs(unordered_map<string, multiset<string>>& m, string s, vector<string>& res) {
while (m[s].size()) {
string t = *m[s].begin();
m[s].erase(m[s].begin());
dfs(m, t, res);
}
res.push_back(s);
}
};

tickets = [["JFK", "KUL"], ["JFK", "NRT"], ["MRT", "JFK"]]

JFK -> KUL, NRT

NRT -> JFK

class Solution {
public:
vector<string> findItinerary(vector<pair<string, string>> tickets) {
vector<string> res;
stack<string> st{{"JFK"}};
unordered_map<string, multiset<string>> m;
for (auto t : tickets) {
m[t.first].insert(t.second);
}
while (!st.empty()) {
string t = st.top();
if (m[t].empty()) {
res.insert(res.begin(), t);
st.pop();
} else {
st.push(*m[t].begin());
m[t].erase(m[t].begin());
}
}
return res;
}
};

Course Schedule

Course Schedule II

https://leetcode.com/problems/reconstruct-itinerary/

https://discuss.leetcode.com/topic/36370/short-ruby-python-java-c

https://discuss.leetcode.com/topic/36721/short-c-dfs-iterative-44ms-solution-with-explanation-no-recursive-calls-no-backtracking

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2016-02-05 15:22  Grandyang  阅读(...)  评论(... 编辑 收藏