合并两个链表【LC1669】

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.

• 思路：遍历链表找到list1中的第$a-1$个节点和第$b+1$个节点，然后将第$a-1$个节点指向list2链表的初始节点，list2链表的尾节点指向list1中的第$b+1$个节点

• 实现

  /**
   * 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 dummyNode = new ListNode(-1, list1);
          ListNode pre = dummyNode;
          ListNode cur = dummyNode.next;
          int idx = 0;
          while (idx != a){
              pre = cur;
              cur = cur.next;  
              idx++;          
          }
          while (idx != b){
              cur = cur.next;
              idx++;
          }
          pre.next = list2;
          while (list2.next != null){
              list2 = list2.next;
          }
  
          list2.next = cur.next;
          cur.next = null;
          return list1;       
      }
  }
  

  • 复杂度分析

    • 时间复杂度：$O(n)$
    • 空间复杂度：$O(1)$