Added Code for Find Intersection of 2 Linked List [Problem 160 of LeetCode]

Author: Yashi11

Linked List/intersection_of_two_linked_lists.cpp

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+/*
+
+	Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null.
+
+
+	Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
+	Output: Intersected at '8'
+	
+	Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
+	Output: Intersected at '2'
+
+	Constraints:
+
+	> The number of nodes of listA is in the m.
+	> The number of nodes of listB is in the n.
+	> 1 <= m, n <= 3 * 104
+	> 1 <= Node.val <= 105
+	> 0 <= skipA < m
+	> 0 <= skipB < n
+	> intersectVal is 0 if listA and listB do not intersect.
+	> intersectVal == listA[skipA] == listB[skipB] if listA and listB intersect.
+
+*/
+
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
+        // Define basic variables needed
+      int len1=0;
+      int len2=0;
+      ListNode* tempA=headA;
+      ListNode* tempB=headB;
+
+      // Calculate the length of both Linked Lists and store in len1 and len2
+      while(tempA!=NULL)
+      {
+          len1++;
+          tempA=tempA->next;
+      } 
+      while(tempB!=NULL)
+      {
+          len2++;
+          tempB=tempB->next;
+      } 
+
+      // Here, we assume that length of Linked-List A is less than or equal
+        // to that of Linked List B
+      if(len1>len2){
+          swap(headA,headB);
+      }
+
+      // Re-initialize variables
+      tempA=headA;
+      tempB=headB;
+      int n=abs(len2-len1);
+      while(n--)
+      {
+          tempB=tempB->next;
+      }
+
+      // Finally, Find the Intersection Node
+      while(tempA!=tempB)
+      {
+          tempA=tempA->next;
+          tempB=tempB->next;
+      } 
+
+      // Return the final answer
+      return tempA;
+    }
+};