Added C solutions

Author: javadev

README.md

@@ -0,0 +1,66 @@
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+
+## 1\. Two Sum
+
+Easy
+
+Given an array of integers `nums` and an integer `target`, return _indices of the two numbers such that they add up to `target`_.
+
+You may assume that each input would have **_exactly_ one solution**, and you may not use the _same_ element twice.
+
+You can return the answer in any order.
+
+**Example 1:**
+
+**Input:** nums = [2,7,11,15], target = 9
+
+**Output:** [0,1]
+
+**Output:** Because nums[0] + nums[1] == 9, we return [0, 1]. 
+
+**Example 2:**
+
+**Input:** nums = [3,2,4], target = 6
+
+**Output:** [1,2] 
+
+**Example 3:**
+
+**Input:** nums = [3,3], target = 6
+
+**Output:** [0,1] 
+
+**Constraints:**
+
+*   <code>2 <= nums.length <= 10<sup>4</sup></code>
+*   <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>
+*   <code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code>
+*   **Only one valid answer exists.**
+
+**Follow-up:** Can you come up with an algorithm that is less than <code>O(n<sup>2</sup>) </code>time complexity?
+
+## Solution
+
+```c
+/**
+ * Note: The returned array must be malloced, assume caller calls free().
+ */
+int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
+    returnSize[0] = 2;
+    int* output = (int*)malloc(sizeof(int) * 2);
+
+    for (int offset = 1; offset < numsSize; offset++) {
+        int i = 0;
+        while (i + offset < numsSize) {
+            if (nums[i] + nums[i + offset] == target) {
+                output[0] = i;
+                output[1] = i + offset;
+                return output;
+            }
+            i++;
+        }
+    }
+    return (void*)0;
+}
+```