<p>You are given two linked lists: <code>list1</code> and <code>list2</code> of sizes <code>n</code> and <code>m</code> respectively.</p>

<p>Remove <code>list1</code>&#39;s nodes from the <code>a<sup>th</sup></code> node to the <code>b<sup>th</sup></code> node, and put <code>list2</code> in their place.</p>

<p>The blue edges and nodes in the following figure indicate the result:</p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/05/fig1.png" style="height: 130px; width: 504px;" />
<p><em>Build the result list and return its head.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/03/01/ll.png" style="width: 609px; height: 210px;" />
<pre>
<strong>Input:</strong> list1 = [10,1,13,6,9,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
<strong>Output:</strong> [10,1,13,1000000,1000001,1000002,5]
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/05/merge_linked_list_ex2.png" style="width: 463px; height: 140px;" />
<pre>
<strong>Input:</strong> list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
<strong>Output:</strong> [0,1,1000000,1000001,1000002,1000003,1000004,6]
<strong>Explanation:</strong> The blue edges and nodes in the above figure indicate the result.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= list1.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= a &lt;= b &lt; list1.length - 1</code></li>
	<li><code>1 &lt;= list2.length &lt;= 10<sup>4</sup></code></li>
</ul>

/**
 * 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 *p1=list1,*p3=list2;
        for(int i=1;i<a;i++)
        {
            p1=p1->next;
        }
        ListNode * start=p1,* end=p1->next;
        for(int i=a;i<=b;i++)
        {
            ListNode *temp=end;
            end=end->next;
            delete temp;
        }
        p1->next=list2;
        while(p3->next)
        {
            p3=p3->next;
        }
        
        p1->next=list2;
        p3->next=end;
        return list1;

    }
};
99800