
Problem Statement
In this task, you have two singly linked lists, list1
and list2
, where the former has size n
and the latter size m
. The goal is to modify list1
by removing a certain segment of its nodes, specifically from the ath
node to the bth
node (both inclusive), and then insert all the nodes of list2
into list1
at the position where the first removed node was. The process transforms list1
by combining parts of it with all of list2
. Post operation, the method should return the head of the updated linked list. The problem asks to ensure this merged linked list reflects the desired sequence of nodes as specified by the indices a
and b
, and the structure of list2
.
Examples
Example 1
Input:
list1 = [10,1,13,6,9,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
Output:
[10,1,13,1000000,1000001,1000002,5]
Explanation:
We remove the nodes 3 and 4 and put the entire list2 in their place. The blue edges and nodes in the above figure indicate the result.
Example 2
Input:
list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
Output:
[0,1,1000000,1000001,1000002,1000003,1000004,6]
Explanation:
The blue edges and nodes in the above figure indicate the result.
Constraints
3 <= list1.length <= 104
1 <= a <= b < list1.length - 1
1 <= list2.length <= 104
Approach and Intuition
To achieve the merging of list2
into list1
at specified positions, follow these systematic steps:
- Identify the nodes just before and just after the segment
[a, b]
inlist1
. Let's call thesepre_a
andpost_b
. - Traverse
list1
to findpre_a
(the node right beforea
). Ifa
is 1,pre_a
will be a new dummy head. - Similarly, find
post_b
(the node right afterb
). - The next step involves disconnecting the segment from
a
tob
:- Set the next pointer of
pre_a
(if it exists) to the head oflist2
.
- Set the next pointer of
- Traverse to the end of
list2
to connect list2’s last node topost_b
. This action effectively insertslist2
. - If
a
equals 1, update the head of the list to be the head oflist2
.
Taking into account the constraints ensures that the values for a
and b
are within valid ranges, and that both linked lists are of permissible lengths. Maintaining the integrity of the linked list while making the insertions and deletions is crucial for the correct manipulation of node pointers. Make sure to appropriately handle cases where list2
might be empty or exceptionally large compared to list1
. Also, consider edge cases such as inserting at the very beginning or the end of the list. This approach combines traversal, pointer manipulation, and requires careful attention to node connections for achieving the desired linked list transformations.
Solutions
- C++
- Java
- Python
class Solution {
public:
ListNode* integrateNodes(ListNode* list1, int startIdx, int endIdx, ListNode* list2) {
return integrate(list1, startIdx, endIdx, list2, 0, nullptr, list1);
}
private:
ListNode* integrate(ListNode* list1, int startIdx, int endIdx, ListNode* list2,
int currentIndex, ListNode* lastNode, ListNode* currentNode) {
if (currentIndex == startIdx - 1) {
lastNode = currentNode;
}
if (currentIndex == endIdx) {
lastNode->next = list2;
ListNode* tail = getTail(list2);
tail->next = currentNode->next;
currentNode->next = nullptr;
return list1;
}
return integrate(list1, startIdx, endIdx, list2, currentIndex + 1, lastNode, currentNode->next);
}
ListNode* getTail(ListNode* node) {
while (node->next != nullptr) {
node = node->next;
}
return node;
}
};
In the given C++ code, the task is to merge two singly linked lists such that one list is inserted into another between two specific indices. The code defined inside the Solution class features a public member function integrateNodes
and a private member function integrate
.
Code Explanation:
The
integrateNodes
function serves as the main entry point. It calls theintegrate
function, beginning the merging process by passing the initial linked list (list1
), the indices between whichlist2
should be integrated (startIdx
andendIdx
), and the head of the second linked list (list2
).The
integrate
function recurses through the nodes oflist1
. It uses the parameters:currentIndex
to keep track of the current node index inlist1
.lastNode
to remember the node just before thestartIdx
.currentNode
as the current node in the traversal oflist1
.
When
currentIndex
equalsstartIdx - 1
, the function updateslastNode
tocurrentNode
, marking the node after whichlist2
should be linked.As
currentIndex
reachesendIdx
, the function performs the splice:lastNode->next
is updated to point tolist2
,tail
(retrieved using the helper functiongetTail
) is the last node oflist2
after which the function connects the part oflist1
that followsendIdx
.
The helper function
getTail
iterates throughlist2
to find its tail node.
**Executing this code will result in list1
being modified in place. The section from startIdx
to endIdx
of list1
is replaced by all nodes in list2
. Following the integration, nodes from endIdx + 1
onward in list1
will continue after the nodes from list2
. This approach uses recursion to efficiently manage the index counting and node traversal, ensuring a seamless integration without managing multiple loops or additional data structures.
class Solution {
public ListNode mergeTwo(ListNode list1, int startIdx, int endIdx, ListNode list2) {
// Initiate merge helper with initial values
return mergeHelper(list1, startIdx, endIdx, list2, 0, null, list1);
}
private ListNode mergeHelper(ListNode list1, int startIdx, int endIdx, ListNode list2,
int currentIdx, ListNode prevNode, ListNode currentNode) {
// Set prevNode at the node before startIdx
if (currentIdx == startIdx - 1) {
prevNode = currentNode;
}
// Recursive base case
if (currentIdx == endIdx) {
// Connect prevNode to list2's head
prevNode.next = list2;
// Search for the last node of list2
ListNode tail = getTail(list2);
tail.next = currentNode.next;
currentNode.next = null;
return list1;
}
return mergeHelper(list1, startIdx, endIdx, list2, currentIdx + 1, prevNode, currentNode.next);
}
private ListNode getTail(ListNode node) {
// Traverse to the end of the list
if (node.next == null) {
return node;
}
return getTail(node.next);
}
}
The given Java solution involves merging one linked list into another between specified positions. The code defines a Solution
class with the method mergeTwo
, which integrates list2
into list1
starting just after the node at index startIdx
and ending at the node at index endIdx
. The integration is done using a helper method, mergeHelper
, that uses recursion to traverse list1
and insert list2
.
- The method
mergeTwo
callsmergeHelper
with initial values representing the current state of the traversal. mergeHelper
traverseslist1
recursively. AtstartIdx - 1
, it records the node just before wherelist2
should be inserted.- When the traversal reaches
endIdx
, it links the last node beforestartIdx
(prevNode
) to the head oflist2
. - The method then continues to find the tail of
list2
using another helper method,getTail
, which recursively reaches the last node oflist2
. - This tail node is then connected to the node following
endIdx
inlist1
, thus linkinglist2
correctly in betweenlist1
. - Finally, by connecting the nodes and handling the recursion base cases properly, the method ensures
list1
is updated and returned withlist2
merged in between the specified indices.
To summarize, this implementation effectively merges two linked lists by manipulating node connections and utilizes recursion for node traversal and merging, making sure the list integrity is maintained throughout the process. Ensure to verify edge cases, like overlapping indices or null lists, for a robust application.
class Solution:
def spliceLists(self, first_list: ListNode, left: int, right: int, second_list: ListNode) -> ListNode:
def get_tail(node: ListNode) -> ListNode:
if node.next is None:
return node
return get_tail(node.next)
def integrate(index, precursor, terminal):
if index == left - 1:
precursor = terminal
if index == right:
precursor.next = second_list
list2_tail = get_tail(second_list)
list2_tail.next = terminal.next
terminal.next = None
return first_list
return integrate(index + 1, precursor, terminal.next)
return integrate(0, None, first_list)
The provided Python code defines a class Solution
with a method spliceLists
that merges one linked list into another at specified positions. The method takes four parameters: first_list
, left
, right
, and second_list
. first_list
is the primary linked list where second_list
will be inserted between the indices left
and right
.
The method uses two helper functions:
get_tail
: Recurses through the list to find and return the tail (the last node) of the list.integrate
: Recursive function handling navigation of thefirst_list
and splicing insecond_list
. It adjusts pointers to mergesecond_list
intofirst_list
between nodes at positionsleft
andright
.
Steps in the process:
- Call
integrate
with initial parameters, starting from the head offirst_list
. - Traverse
first_list
to find the node just before positionleft
(pre-modification point). - Attach the start of
second_list
to this node. - Find the tail of
second_list
usingget_tail
. - Attach the node following position
right
infirst_list
to the tail ofsecond_list
, effectively insertingsecond_list
betweenleft
andright
. - Return the head of the modified
first_list
.
The approach ensures that the insertion of second list is seamless, controlled through recursive traversal and pointer reassignment to maintain list integrity. Handling edge cases, like list bounds, would be implicit in the logic through careful index management and null handling. Ensuring efficient traversal and proper tail connection is crucial for maintaining linked list structure.
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