Added to Linked-List

Author: sr-sweta

Data Structures/Linked Lists/Singly Linked List/Intersection_Of_Two_LinkedList.cpp

@@ -0,0 +1,78 @@
+/*
+Src : LeetCode
+--------------
+
+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.
+For example, the following two linked lists begin to intersect at node c1:
+
+ A:               a1 --- a2
+                		  \
+                		   c1 --- c2 --- c3
+                		  / 
+ B:        b1 --- b2 --- b3
+
+It is guaranteed that there are no cycles anywhere in the entire linked structure.
+Note that the linked lists must retain their original structure after the function returns.
+
+Example: 
+
+ A:               4 --- 1
+                		  \
+                		   8 --- 4 --- 5
+                		  / 
+ B:        5 --- 6 --- 1
+
+Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
+Output: Intersected at '8'
+Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect).
+From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
+
+
+Constraints:
+
+#  The number of nodes of listA is in the m.
+#  The number of nodes of listB is in the n.
+#  0 <= m, n <= 3 * 104
+#  1 <= Node.val <= 105
+#  0 <= skipA <= m
+#  0 <= skipB <= n
+#  intersectVal is 0 if listA and listB do not intersect.
+3  intersectVal == listA[skipA + 1] == listB[skipB + 1] if listA and listB intersect.
+
+
+Follow up: Could you write a solution that runs in O(n) time and use only O(1) memory?
+
+*/
+
+/**
+ * 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) {
+        ListNode *p1 = headA;
+    	ListNode *p2 = headB;
+        
+	    if (p1 == NULL || p2 == NULL) return NULL;
+	
+	    while (p1 != NULL && p2 != NULL && p1 != p2) {
+	        p1 = p1->next;
+	        p2 = p2->next;
+	        if (p1 == p2) return p1;
+	        if (p1 == NULL) p1 = headB;
+	        if (p2 == NULL) p2 = headA;
+	    }
+	        
+	    return p1;
+    }
+};
+
+
+//CODE
+//Time Complexity = O(n) 
+//Space Complexity = O(1)