1669. Merge In Between Linked Lists | LeetCode Solution

 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;

    }

};