LeetCode-探索-初级算法-链表-4. 合并两个有序链表(个人做题记录，不是习题讲解)

LeetCode-探索-初级算法-链表-4. 合并两个有序链表(个人做题记录，不是习题讲解)

LeetCode探索-初级算法：https://leetcode-cn.com/explore/interview/card/top-interview-questions-easy/

4. 合并两个有序链表

• 语言：java

• 思路：新建一个ListNode作为head，然后判断链表a和链表b的数值大小，每次往head后面添加子节点

• 代码(1ms)：

  /**
   * Definition for singly-linked list.
   * public class ListNode {
   *     int val;
   *     ListNode next;
   *     ListNode(int x) { val = x; }
   * }
   */
  class Solution {
      public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
          ListNode a = l1;
          ListNode b = l2;
          ListNode head = new ListNode(0);
          ListNode res = head;
          while(a!=null&&b!=null){
              if(a.val<=b.val){
                  head.next = a;
                  a = a.next;
                  head = head.next;
              }else{
                  head.next = b;
                  b = b.next;
                  head = head.next;
              }
          }
          if(a!=null)
              head.next = a;
          else if(b!=null)
              head.next = b;
          return res.next;
      }
  }
  

• 参考代码(0ms)递归:

  ​ 直接在原本的ab两个链表上操作，所以最后先遍历到null的返回另外一个链表；每次递归判断是要a后面添加b还是b后面添加a，如果a后面添加b，那么久返回a；

  class Solution {
      public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
          if (l1 == null) {
              return l2;
          } else if (l2 == null) {
              return l1;
          } else if (l1.val < l2.val) {
              l1.next = mergeTwoLists(l1.next, l2);
              return l1;
          } else {
              l2.next = mergeTwoLists(l1, l2.next);
              return l2;
          }
      }
  }
  

• 参考后重写：

  /**
   * Definition for singly-linked list.
   * public class ListNode {
   *     int val;
   *     ListNode next;
   *     ListNode(int x) { val = x; }
   * }
   */
  class Solution {
      public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
          if(l1==null)
              return l2;
          else if(l2==null)
              return l1;
          else{
              if(l1.val<=l2.val){
                  l1.next = mergeTwoLists(l1.next,l2);
                  return l1;
              }else{
                  l2.next = mergeTwoLists(l1,l2.next);
                  return l2;
              }
          }
      }
  }