New Problem Solution "Reconstruct Itinerary"

README.md

@@ -54,6 +54,7 @@ LeetCode
 |338|[Counting Bits](https://leetcode.com/problems/counting-bits/) | [C++](./algorithms/cpp/countingBits/CountingBits.cpp)|Medium|
 |337|[House Robber III](https://leetcode.com/problems/house-robber-iii/) | [C++](./algorithms/cpp/houseRobber/houseRobberIII.cpp)|Medium|
 |334|[Increasing Triplet Subsequence](https://leetcode.com/problems/increasing-triplet-subsequence/) | [C++](./algorithms/cpp/increasingTripletSubsequence/increasingTripletSubsequence.cpp)|Medium|
+|332|[Reconstruct Itinerary](https://leetcode.com/problems/reconstruct-itinerary/) | [C++](./algorithms/cpp/reconstructItinerary/ReconstructItinerary.cpp)|Medium|
 |331|[Verify Preorder Serialization of a Binary Tree](https://leetcode.com/problems/verify-preorder-serialization-of-a-binary-tree/) | [C++](./algorithms/cpp/verifyPreorderSerializationOfABinaryTree/VerifyPreorderSerializationOfABinaryTree.cpp)|Medium|
 |330|[Patching Array](https://leetcode.com/problems/patching-array/) | [C++](./algorithms/cpp/patchingArray/PatchingArray.cpp)|Medium|
 |329|[Longest Increasing Path in a Matrix](https://leetcode.com/problems/longest-increasing-path-in-a-matrix/) | [C++](./algorithms/cpp/longestIncreasingPathInAMatrix/LongestIncreasingPathInAMatrix.cpp)|Medium|

algorithms/cpp/reconstructItinerary/ReconstructItinerary.cpp

@@ -0,0 +1,78 @@
+// Source : https://leetcode.com/problems/reconstruct-itinerary/
+// Author : Hao Chen
+// Date   : 2017-01-06
+
+/*************************************************************************************** 
+ *
+ * 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:
+ *     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.
+ ***************************************************************************************/
+
+/*
+ This problem's description really confuse me.
+ 
+ for examples:
+    1) [["JFK", "PEK"], ["JFK", "SHA"], ["SHA", "JFK"]], which has two itineraries: 
+       a) JFK -> PEK, 
+       b) JFK -> SHA -> JFK -> PEK
+       The a) is smaller than b), because PEK < SHA, however the b) is correct answer.
+       So, it means we need use all of tickets.
+       
+    2) [["JFK", "PEK"], ["JFK", "SHA"]], which also has two itineraries:
+       a) JFK -> PEK
+       b) JFK -> SHA
+       for my understanding, the JFK -> SHA is the correct one, 
+       however, the correct answer is JFK -> SHA -> PEK.
+       I don't understand, why the correct answer is not JFK -> PEK -> SHA
+       That really does not make sense to me.
+       
+ All right, we need assume all of the tickets can be connected in one itinerary.
+ Then, it's easy to have a DFS algorithm.
+*/
+
+
+class Solution {
+public:
+    //DFS
+    void travel(string& start, unordered_map<string, multiset<string>>& map, vector<string>& result) {
+        while (map[start].size() > 0 ) {
+            string next = *(map[start].begin());
+            map[start].erase(map[start].begin());
+            travel(next, map, result);
+        }
+        result.insert(result.begin(), start);
+    }
+    
+    vector<string> findItinerary(vector<pair<string, string>> tickets) {
+        unordered_map<string, multiset<string>> map;
+        for(auto t : tickets) {
+            map[t.first].insert(t.second);
+        }
+        vector<string> result;
+        string start = "JFK";
+        travel(start, map, result);
+        return result;
+    }
+};