leetcode 21. Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

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# Definition for singly-linked list. # class ListNode(object): #     def __init__(self, x): #         self.val = x #         self.next = None   class Solution(object):     def mergeTwoLists(self, l1, l2):         """         :type l1: ListNode         :type l2: ListNode         :rtype: ListNode         """         dummy = cur = ListNode(None)         while l1 and l2:             if l1.val <= l2.val:                 cur.next = ListNode(l1.val)                 l1 = l1.next             else:                 cur.next = ListNode(l2.val)                 l2 = l2.next             cur = cur.next         l = l1 or l2         while l:             cur.next = ListNode(l.val)             l = l.next             cur = cur.next         return dummy.next

可以稍微少几行代码：

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# Definition for singly-linked list. # class ListNode(object): #     def __init__(self, x): #         self.val = x #         self.next = None   class Solution(object):     def mergeTwoLists(self, l1, l2):         """         :type l1: ListNode         :type l2: ListNode         :rtype: ListNode         """         dummy = cur = ListNode(None)         while l1 or l2:             if (l1 and l2 and l1.val <= l2.val) or l2 is None:                 cur.next = ListNode(l1.val)                 l1 = l1.next             elif (l1 and l2 and l1.val > l2.val) or l1 is None:                 cur.next = ListNode(l2.val)                 l2 = l2.next             cur = cur.next                return dummy.next

 

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# Definition for singly-linked list. # class ListNode(object): #     def __init__(self, x): #         self.val = x #         self.next = None   class Solution(object):     def mergeTwoLists(self, l1, l2):         """         :type l1: ListNode         :type l2: ListNode         :rtype: ListNode         """         if (l1 and l2 and l1.val <= l2.val) or (l1 and not l2):             node = ListNode(l1.val)             node.next = self.mergeTwoLists(l1.next, l2)             return node         elif (l1 and l2 and l1.val > l2.val) or (l2 and not l1):             node = ListNode(l2.val)             node.next = self.mergeTwoLists(l1, l2.next)             return node         else:                            return None

还有偷懒的解法：

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class Solution { public:     ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {         ListNode dummy(INT_MIN);         ListNode *tail = &dummy;                   while (l1 && l2) {             if (l1->val < l2->val) {                 tail->next = l1;                 l1 = l1->next;             } else {                 tail->next = l2;                 l2 = l2->next;             }             tail = tail->next;         }           tail->next = l1 ? l1 : l2;         return dummy.next;     } };

 看到其他人的解法，如果允许修改原有list的话，还是可以的：

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# Definition for singly-linked list. # class ListNode(object): #     def __init__(self, x): #         self.val = x #         self.next = None   class Solution(object):     def mergeTwoLists(self, l1, l2):         """         :type l1: ListNode         :type l2: ListNode         :rtype: ListNode         """         dummy = cur = ListNode(None)         while l1 and l2:             if l1.val < l2.val:                 cur.next = l1                 l1 = l1.next             else:                 cur.next = l2                 l2 = l2.next             cur = cur.next         cur.next = l1 or l2         return dummy.next

 对应的递归解法：

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# Definition for singly-linked list. # class ListNode(object): #     def __init__(self, x): #         self.val = x #         self.next = None   class Solution(object):     def mergeTwoLists(self, l1, l2):         """         :type l1: ListNode         :type l2: ListNode         :rtype: ListNode         """         if l1 is None: return l2         if l2 is 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