New Problem Solution -"Restore the Array From Adjacent Pairs"

README.md

@@ -54,6 +54,7 @@ LeetCode
 |1750|[Minimum Length of String After Deleting Similar Ends](https://leetcode.com/problems/minimum-length-of-string-after-deleting-similar-ends/) | [C++](./algorithms/cpp/minimumLengthOfStringAfterDeletingSimilarEnds/MinimumLengthOfStringAfterDeletingSimilarEnds.cpp)|Medium|
 |1749|[Maximum Absolute Sum of Any Subarray](https://leetcode.com/problems/maximum-absolute-sum-of-any-subarray/) | [C++](./algorithms/cpp/maximumAbsoluteSumOfAnySubarray/MaximumAbsoluteSumOfAnySubarray.cpp)|Medium|
 |1748|[Sum of Unique Elements](https://leetcode.com/problems/sum-of-unique-elements/) | [C++](./algorithms/cpp/sumOfUniqueElements/SumOfUniqueElements.cpp)|Easy|
+|1743|[Restore the Array From Adjacent Pairs](https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/) | [C++](./algorithms/cpp/restoreTheArrayFromAdjacentPairs/RestoreTheArrayFromAdjacentPairs.cpp)|Medium|
 |1742|[Maximum Number of Balls in a Box](https://leetcode.com/problems/maximum-number-of-balls-in-a-box/) | [C++](./algorithms/cpp/maximumNumberOfBallsInABox/MaximumNumberOfBallsInABox.cpp)|Easy|
 |1736|[Latest Time by Replacing Hidden Digits](https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/) | [C++](./algorithms/cpp/latestTimeByReplacingHiddenDigits/LatestTimeByReplacingHiddenDigits.cpp)|Easy|
 |1734|[Decode XORed Permutation](https://leetcode.com/problems/decode-xored-permutation/) | [C++](./algorithms/cpp/decodeXORedPermutation/DecodeXoredPermutation.cpp)|Medium|

algorithms/cpp/restoreTheArrayFromAdjacentPairs/RestoreTheArrayFromAdjacentPairs.cpp

@@ -0,0 +1,81 @@
+// Source : https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/
+// Author : Hao Chen
+// Date   : 2021-03-28
+
+/***************************************************************************************************** 
+ *
+ * There is an integer array nums that consists of n unique elements, but you have forgotten it. 
+ * However, you do remember every pair of adjacent elements in nums.
+ * 
+ * You are given a 2D integer array adjacentPairs of size n - 1 where each adjacentPairs[i] = [ui, vi] 
+ * indicates that the elements ui and vi are adjacent in nums.
+ * 
+ * It is guaranteed that every adjacent pair of elements nums[i] and nums[i+1] will exist in 
+ * adjacentPairs, either as [nums[i], nums[i+1]] or [nums[i+1], nums[i]]. The pairs can appear in any 
+ * order.
+ * 
+ * Return the original array nums. If there are multiple solutions, return any of them.
+ * 
+ * Example 1:
+ * 
+ * Input: adjacentPairs = [[2,1],[3,4],[3,2]]
+ * Output: [1,2,3,4]
+ * Explanation: This array has all its adjacent pairs in adjacentPairs.
+ * Notice that adjacentPairs[i] may not be in left-to-right order.
+ * 
+ * Example 2:
+ * 
+ * Input: adjacentPairs = [[4,-2],[1,4],[-3,1]]
+ * Output: [-2,4,1,-3]
+ * Explanation: There can be negative numbers.
+ * Another solution is [-3,1,4,-2], which would also be accepted.
+ * 
+ * Example 3:
+ * 
+ * Input: adjacentPairs = [[100000,-100000]]
+ * Output: [100000,-100000]
+ * 
+ * Constraints:
+ * 
+ * 	nums.length == n
+ * 	adjacentPairs.length == n - 1
+ * 	adjacentPairs[i].length == 2
+ * 	2 <= n <= 10^5
+ * 	-10^5 <= nums[i], ui, vi <= 10^5
+ * 	There exists some nums that has adjacentPairs as its pairs.
+ ******************************************************************************************************/
+
+class Solution {
+public:
+    vector<int> restoreArray(vector<vector<int>>& adjacentPairs) {
+        // only two numbers have one neighbour
+        // start from one of them to travel all number.
+        
+        unordered_map<int, vector<int>> dict;
+        for(auto& pair : adjacentPairs) {
+            dict[pair[0]].push_back(pair[1]);
+            dict[pair[1]].push_back(pair[0]);
+        }
+        
+        int end[2]; int i=0;
+        for (auto& [key, pair] : dict) {
+            if(pair.size()==1) end[i++] = key;
+            if (i>1) break;
+        }
+        //cout << "start=" << end[0] <<", end=" << end[1] << endl;
+        vector<int> result(1, end[0]);
+        int start = end[0];
+        int prev = -1;
+        while ( start != end[1] ) {
+            auto& v = dict[start];
+            for(int i= 0; i< v.size(); i++) {
+                if (v[i] == prev) continue;
+                result.push_back(v[i]);
+                prev = start;
+                start = v[i];
+                break;
+            }
+        }
+        return result;
+    }
+};