# 1669. Merge In Between Linked Lists

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 incidate 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

## Solutions

### 1. straight forward

```cpp
/**
 * 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) {
        a++; b++;
        ListNode dummy(0, list1), * prev1, * prev2;
        prev1 = prev2 = &dummy;

        while (--a > 0)
            prev1 = prev1->next;
        while (b-- > 0)
            prev2 = prev2->next;
        prev1->next = list2;
        while (prev1->next)
            prev1 = prev1->next;
        prev1->next = prev2->next;

        return dummy.next;
    }
};
```