How do you find the intersection of two linked lists in C++?
How do you find the intersection of two linked lists in C++?
To find the intersection point, we will traverse both the linked lists till we find that they are equally pointing to the same value. At some point, the pointer to the next node of the linked list will be the same. Thus we will return the value of that point.
Is it a possible to produce union of two linked lists?
Union of two linked lists can be found by using merging the lists in a sorted manner. The intersection of the two lists can be found by only taking common elements while merging the two lists.
How do you find the sum of two linked lists using stack?
Algorithm: The formal steps of this algorithm are as following:
- Create stack ‘s1’ by pushing all node values of the first linked list to a stack.
- Create stack ‘s2’ by pushing all node values of the second linked list to a stack.
- Create an empty stack ‘s3’ to store the result of addition.
- Initialize sum and carry to 0.
How to solve the intersection of two linked lists?
Input: First linked list: 1->2->3->4->5 Second linked list be 2->3->4, Output: 2->3->4 The elements 2, 3, 4 are common in both the list so they appear in the intersection list. Recommended: Please solve it on “ PRACTICE ” first, before moving on to the solution.
How to get the point where two linked list Merge?
Write a program to get the point where two linked list merge. Above diagram shows an example with two linked list having 15 as intersection point. Recommended: Please solve it on “ PRACTICE ” first, before moving on to the solution.
How to calculate time complexity of two linked lists?
Time Complexity: O (m+n) where m and n are number of nodes in first and second linked lists respectively. Only one traversal of the lists are needed. Auxiliary Space: O (min (m, n)). The output list can store at most min (m,n) nodes .
Are there any cycles in a linked list?
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.