21. 合并两个有序链表

题目来源：21. 合并两个有序链表

将两个升序链表合并为一个新的 升序 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 1：

输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]

示例 2：

输入：l1 = [], l2 = []
输出：[]

示例 3：

输入：l1 = [], l2 = [0]
输出：[0]

 

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var mergeTwoLists = function(l1, l2) {
  let fir = l1;
  let sec = l2;
  let res = null;
  let tail = null;
  while(fir != null && sec != null){
    let cur = null;
    if(fir.val <= sec.val){
      cur = fir;
      fir = fir.next;
    }else{
      cur = sec;
      sec = sec.next;
    }
    if(res == null){
      res = cur;
      tail = res;
    }else{
      tail.next = cur;
      tail = cur;
    }
  }
  let next = (fir == null)?sec:fir;
  if(tail == null){
    res = next;
  }else{
    tail.next = next;
  }
  return res;
};

提示：

• 两个链表的节点数目范围是 [0, 50]
• -100 <= Node.val <= 100
• l1 和 l2 均按 非递减顺序 排列

Python3

# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
      if l1 is None:
        return l2
      elif l2 is None:
        return l1
      elif l1.val < l2.val:
        l1.next = self.mergeTwoLists(l1.next, l2)
        return l1
      else:
        l2.next = self.mergeTwoLists(l1, l2.next)
        return l2    

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
      preHead = ListNode(-1)
      pre = preHead
      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

      pre.next = l1 if l1 is not None else l2

      return preHead.next