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

Solution: List Operations

Find the following nodes:
1. previous node to the a-th node: prev_a
2. the b-th node: node_b
3. tail node of list2: tail2

prev_a->next = list2
tail2->next = node_b

return list1

Time complexity: O(m+n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {
    ListNode dummy(0, list1);
    ListNode* prev_a = &dummy;
    for (int i = 0; i < a; ++i) prev_a = prev_a->next;
    ListNode* node_b = prev_a->next;
    for (int i = a; i <= b; ++i) node_b = node_b->next;
    ListNode* tail2 = list2;
    while (tail2->next) tail2 = tail2->next;
    
    prev_a->next = list2;    
    tail2->next = node_b;
    return list1;
  }
};

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