add c++ solution for 1

fishercoder1534 · fishercoder1534 · commit 8a74f49036eb · 2019-03-12T21:34:30.000-07:00
diff --git a/README.md b/README.md
@@ -764,7 +764,7 @@ Your ideas/fixes/algorithms are more than welcome!
 |4|[Median of Two Sorted Arrays](https://leetcode.com/problems/median-of-two-sorted-arrays/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_4.java) | ? | ? | |Hard | Divide and Conquer
 |3|[Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_3.java) | O(n) | O(k) | |Medium | HashMap, Sliding Window
 |2|[Add Two Numbers](https://leetcode.com/problems/add-two-numbers/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_2.java) | O(max(m,n)) | O(1) | |Medium | LinkedList
-|1|[Two Sum](https://leetcode.com/problems/two-sum/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1.java)| O(n)| O(n) |[:tv:](https://www.youtube.com/watch?v=kPXOr6pW8KM&t=)|Easy| HashMap 
+|1|[Two Sum](https://leetcode.com/problems/two-sum/)|[Java](../master/src/main/java/com/fishercoder/solutions/_1.java), [C++](../master/cpp/_1.cpp)| O(n)| O(n) |[:tv:](https://www.youtube.com/watch?v=kPXOr6pW8KM&t=)|Easy| HashMap 
 
 ## Database
 
diff --git a/cpp/_1.cpp b/cpp/_1.cpp
@@ -0,0 +1,35 @@
+/**
+1. Two Sum
+
+Given an array of integers, return indices of the two numbers such that they add up to a specific target.
+
+You may assume that each input would have exactly one solution, and you may not use the same element twice.
+
+Example:
+
+Given nums = [2, 7, 11, 15], target = 9,
+
+Because nums[0] + nums[1] = 2 + 7 = 9,
+return [0, 1].
+
+*/
+class Solution {
+public:
+    vector<int> twoSum(vector<int>& nums, int target) {
+        unordered_map<int, int> m;
+        vector<int> result;
+        for(int i=0; i<nums.size(); i++){
+        // not found the second one
+            if (m.find(nums[i])==m.end() ) {
+            // store the first one poisition into the second one's key
+                m[target - nums[i]] = i;
+            }else {
+            // found the second one
+                result.push_back(m[nums[i]]);
+                result.push_back(i);
+                break;
+            }
+        }
+        return result;
+    }
+};
