1669. Merge In Between Linked Lists | LeetCode Solution
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 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002] Output: [0,1,2,1000000,1000001,1000002,5] 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
/**
* 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) {
int m=a,n=b;
ListNode* ans=list1;
while(a>1){
list1=list1->next;
a--;
}
ListNode* add=ans;
while(b>0){
add=add->next;
b--;
}
list1->next=list2;
while(list1->next!=NULL)list1=list1->next;
list1->next=add->next;
return ans;
}
};
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