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

---

# 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：

# 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

 

# 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：

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：

# 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：

# 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