Question

1669. Merge In Between Linked Lists

中文文档

Description

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 = [10,1,13,6,9,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
Output: [10,1,13,1000000,1000001,1000002,5]
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

Solution 1: Simulation

We can directly simulate the operations described in the problem.

In the implementation, we use two pointers $p$ and $q$, both initially pointing to the head node of list1.

Then we move pointers $p$ and $q$ forward, until pointer $p$ points to the node before the $a$-th node in list1, and pointer $q$ points to the $b$-th node in list1. At this point, we set the next pointer of $p$ to the head node of list2, and set the next pointer of the tail node of list2 to the node pointed to by the next pointer of $q$. This completes the operation required by the problem.

The time complexity is $O(m + n)$, and the space complexity is $O(1)$. Where $m$ and $n$ are the lengths of list1 and list2 respectively.

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, val=0, next=None):
#     self.val = val
#     self.next = next
class Solution:
  def mergeInBetween(
    self, list1: ListNode, a: int, b: int, list2: ListNode
  ) -> ListNode:
    p = q = list1
    for _ in range(a - 1):
      p = p.next
    for _ in range(b):
      q = q.next
    p.next = list2
    while p.next:
      p = p.next
    p.next = q.next
    q.next = None
    return list1

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode() {}
 *   ListNode(int val) { this.val = val; }
 *   ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
  public ListNode mergeInBetween(ListNode list1, int a, int b, ListNode list2) {
    ListNode p = list1, q = list1;
    while (--a > 0) {
      p = p.next;
    }
    while (b-- > 0) {
      q = q.next;
    }
    p.next = list2;
    while (p.next != null) {
      p = p.next;
    }
    p.next = q.next;
    q.next = null;
    return list1;
  }
}

/**
 * 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) {
    auto p = list1, q = list1;
    while (--a) {
      p = p->next;
    }
    while (b--) {
      q = q->next;
    }
    p->next = list2;
    while (p->next) {
      p = p->next;
    }
    p->next = q->next;
    q->next = nullptr;
    return list1;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func mergeInBetween(list1 *ListNode, a int, b int, list2 *ListNode) *ListNode {
  p, q := list1, list1
  for ; a > 1; a-- {
    p = p.Next
  }
  for ; b > 0; b-- {
    q = q.Next
  }
  p.Next = list2
  for p.Next != nil {
    p = p.Next
  }
  p.Next = q.Next
  q.Next = nil
  return list1
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *   val: number
 *   next: ListNode | null
 *   constructor(val?: number, next?: ListNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 *   }
 * }
 */

function mergeInBetween(
  list1: ListNode | null,
  a: number,
  b: number,
  list2: ListNode | null,
): ListNode | null {
  let p = list1;
  let q = list1;
  while (--a > 0) {
    p = p.next;
  }
  while (b-- > 0) {
    q = q.next;
  }
  p.next = list2;
  while (p.next) {
    p = p.next;
  }
  p.next = q.next;
  q.next = null;
  return list1;
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   public int val;
 *   public ListNode next;
 *   public ListNode(int val=0, ListNode next=null) {
 *     this.val = val;
 *     this.next = next;
 *   }
 * }
 */
public class Solution {
  public ListNode MergeInBetween(ListNode list1, int a, int b, ListNode list2) {
    ListNode p = list1, q = list1;
    while (--a > 0) {
      p = p.next;
    }
    while (b-- > 0) {
      q = q.next;
    }
    p.next = list2;
    while (p.next != null) {
      p = p.next;
    }
    p.next = q.next;
    q.next = null;
    return list1;
  }
}