21. 合并两个有序链表


官方链接

https://leetcode-cn.com/problems/merge-two-sorted-lists/

将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 

示例:

输入:1->2->4, 1->3->4
输出:1->1->2->3->4->4

解法一

/**
 * 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 mergeTwoLists(ListNode l1, ListNode l2) {

        if (l1 == null) return l2;
        if (l2 == null) return l1;

        ListNode dummy = new ListNode(-1);
        ListNode pre = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                pre.next = l1;
                l1 = l1.next;
            } else {
                pre.next = l2;
                l2 = l2.next;
            }
            pre = pre.next;
        }

        if (l1 != null) pre.next = l1;
        else pre.next = l2;

        return dummy.next;
    }
}
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
      dummy = ListNode(-1)
      pre = dummy
      while l1 and l2:
        if l1.val < l2.val:
          pre.next = l1
          l1 = l1.next
        else:
          pre.next = l2
          l2 = l2.next
        pre = pre.next

      if l1: 
        pre.next = l1
      else:
        pre.next = l2
      return dummy.next