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;
    }
};

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