Runtime: 101 ms (Top 23.35%) | Memory: 14.3 MB (Top 98.77%)

Author: AnasImloul

scripts/algorithms/I/Intersection of Two Linked Lists/Intersection of Two Linked Lists.cpp

@@ -1,75 +1,76 @@
+// Runtime: 101 ms (Top 23.35%) | Memory: 14.3 MB (Top 98.77%)
 /**
  * Definition for singly-linked list.
  * struct ListNode {
- *     int val;
- *     ListNode *next;
- *     ListNode(int x) : val(x), next(NULL) {}
+ * int val;
+ * ListNode *next;
+ * ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
-    
+
     //function to get length of linked list
     int getLength(ListNode *head)
     {
         //initial length 0
         int l = 0;
-        
+
         while(head != NULL)
         {
             l++;
             head = head->next;
         }
         return l;
     }
-    
+
     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
-        
+
         //ptr1 & ptr2 will move to check nodes are intersecting or not
         //ptr1 will point to the list which have higher length, higher elements (big list)
         //ptr2 will point to small list
         ListNode *ptr1,*ptr2;
-        
+
         //fetching length of both the list,as we want a node that both ptr points simultaneously
         //so if both list have same length of nodes, they travel with same speed, they intersect
         //if diff legths, it'll be difficult for us to catch
         //so by substracting list with higher length with lower length
         //we get same level of length, from which both ptr start travaeling, they may get intersect
         int length1 = getLength(headA);
         int length2 = getLength(headB);
-        
+
         int diff = 0;
-        
+
         //ptr1 points to longer list
         if(length1>length2)
         {
             diff = length1-length2;
             ptr1 = headA;
             ptr2 = headB;
-            
+
         }
         else
         {
             diff = length2 - length1;
             ptr1 = headB;
             ptr2 = headA;
         }
-        
+
         //till diff is zero and we reach our desired posn
         while(diff)
         {
             //incrementing ptr1 so that it can be at a place where both list have same length
-            
+
             ptr1 = ptr1->next;
-            
+
             if(ptr1 == NULL)
             {
                 return NULL;
             }
-            
+
             diff--;
         }
-        
+
         //traverse both pointers together, till any of them gets null
         while((ptr1 != NULL) && (ptr2 != NULL))
         {
@@ -78,14 +79,14 @@ class Solution {
             {
                 return ptr1;
             }
-            
+
             //increment both pointers, till they both be NULL
             ptr1 = ptr1->next;
             ptr2 = ptr2->next;
         }
-        
+
         //if not found any intersaction, return null
         return NULL;
-        
+
      }
 };