Add solution to day 15

05- May/15- Swapping Nodes in a Linked List/15- Swapping Nodes in a Linked List (Ahmed Gamal).cpp

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+// Author: Ahmed Gamal
+
+// this is a simple implementation problem
+// we will use two pointers to point to the kth from the beginning and the kth from the end
+// then we will swap them
+// to get the kth from the end, we will use the size of the list - k + 1
+// to get the size of the list, we will iterate over the list and count the number of nodes
+
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode() : val(0), next(nullptr) {}
+ *     ListNode(int x) : val(x), next(nullptr) {}
+ *     ListNode(int x, ListNode *next) : val(x), next(next) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* swapNodes(ListNode* head, int k) {
+        // size: the size of the list
+        // i: the kth node from the beginning
+        // j: the kth node from the end
+        int size = 0;
+        ListNode *i;
+        auto j = head;
+
+        // we will iterate over the list and count the number of nodes
+        while(j) {
+            size++;
+
+            // if we reached the kth node from the beginning, we will save it in i
+            if(size == k) {
+                i = j;
+            }
+            j = j -> next;
+        }
+
+        // we will get the kth node from the end (size - k + 1)
+        size -= k;
+        j = head;
+        while(size--) {
+            j = j -> next;
+        }
+
+        // we will swap the values of the two nodes
+        swap(i -> val, j -> val);
+
+        // we will return the head of the list
+        return head;
+    }
+};