Description
You are given two linked lists: list1
and list2
of sizes n
and m
respectively.
Remove list1
's nodes from the ath
node to the bth
node, and put list2
in their place.
The blue edges and nodes in the following figure indicate the result:

Build the result list and return its head.
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
Solution
Python3
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:
curr = res = ListNode(-1, list1)
for _ in range(a):
curr = curr.next
prev = curr
for _ in range(b - a + 1):
curr = curr.next
l = list2
while l.next:
l = l.next
l.next = curr.next
prev.next = list2
return res.next
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {
ListNode *curr = new ListNode;
curr->next = list1;
ListNode *res = curr;
for (int i = 0; i < a; i++)
curr = curr->next;
ListNode *prev = curr;
for (int i = 0; i < b-a+1; i++)
curr = curr->next;
ListNode *temp = list2;
while (temp->next)
temp = temp->next;
temp->next = curr->next;
prev->next = list2;
return res->next;
}
};