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。
方法一: 递归 题解
执行用时:56 ms, 在所有 Python3 提交中击败了18.41%的用户 内存消耗:13.7 MB, 在所有 Python3 提交中击败了7.14%的用户 # 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: if l1 and l2: if l1.val > l2.val: l1, l2 = l2, l1 l1.next = self.mergeTwoLists(l1.next, l2) return l1 or l2 """ For Example: input: l1:1->2->4 l2:1->3->4 output: 1->1->2->3->4->4 """ l1 = ListNode(1) l1.next = ListNode(2) l1.next.next = ListNode(4) l2 = ListNode(1) l2.next = ListNode(3) l2.next.next = ListNode(4) solution = Solution() result = solution.mergeTwoLists(l1, l2) print('输出为:%d->%d->%d->%d->%d->%d' % \ (result.val, result.next.val, result.next.next.val, result.next.next.next.val,\ result.next.next.next.next.val, result.next.next.next.next.next.val))方法二: 递循环遍历 题解
执行用时:44 ms, 在所有 Python3 提交中击败了79.61%的用户 内存消耗:13.8 MB, 在所有 Python3 提交中击败了7.14%的用户 # 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(0) move = dummy while l1 and l2: if l1.val <= l2.val: move.next = l1 l1 = l1.next else: move.next = l2 l2 = l2.next move = move.next move.next = l1 if l1 else l2 return dummy.next """ For Example: input: l1:1->2->4 l2:1->3->4 output: 1->1->2->3->4->4 """ l1 = ListNode(1) l1.next = ListNode(2) l1.next.next = ListNode(4) l2 = ListNode(1) l2.next = ListNode(3) l2.next.next = ListNode(4) solution = Solution() result = solution.mergeTwoLists(l1, l2) print('输出为:%d->%d->%d->%d->%d->%d' % \ (result.val, result.next.val, result.next.next.val, result.next.next.next.val,\ result.next.next.next.next.val, result.next.next.next.next.next.val))