题目
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： 输入：1->2->4, 1->3->4 输出：1->1->2->3->4->4
解题思路
# 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: # tempListNode = ListNode() # cur = tempListNode # while l1 != None and l2 != None: # if l1.val < l2.val:#l1塞入新队列 # cur.next = l1 # l1 = l1.next # else:#l2塞入新队列 # cur.next = l2 # l2 = l2.next # cur = cur.next # #剩余队列进行拼接 # if l1 != None:cur.next = l1 # if l2 != None:cur.next = l2 # return tempListNode.next #递归 if l1 == None : return l2 if l2 == None : return l1 if l1.val <= l2.val: l1.next = self.mergeTwoLists(l1.next, l2) return l1 else: l2.next = self.mergeTwoLists(l1, l2.next) return l2 if __name__ == '__main__': #链表样例，注意复制！ #L1 1->2->3->4->5 l1 = ListNode(1,ListNode(2, ListNode(3, ListNode(4, ListNode(5))))) #L2 1->3->4 l2 = ListNode(1, ListNode(3, ListNode(4))) ret = Solution().mergeTwoLists(l1, l2) print(ret.val) print(ret.next.val) print(ret.next.next.val) print(ret.next.next.next.val) print(ret.next.next.next.next.val)
