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。
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]
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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/merge-two-sorted-lists
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
解题思路
保存头指针,移动当前指针,比较大小,最后将还没遍历完的直接接上
# Definition for singly-linked list. # class ListNode(object): # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution(object): def mergeTwoLists(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ cur = ListNode() temp = cur # 保存头指针,移动的是cur指针 while l1 and l2: if l1.val <= l2.val: cur.next = l1 l1 = l1.next elif l1.val > l2.val: cur.next = l2 l2 = l2.next cur = cur.next if l1 is not None: cur.next = l1 elif l2 is not None: cur.next = l2 return temp.next 作者:zcw1234515 链接:https://leetcode-cn.com/problems/merge-two-sorted-lists/solution/21-he-bing-liang-ge-you-xu-lian-biao-by-37zib/ 来源:力扣(LeetCode) 著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。