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.

我的解决方式：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
 

class Solution 
{
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) 
    {
        if(NULL==l1) return l2;
        if(NULL==l2) return l1;
       
         ListNode* head = NULL;
         
         if(l1->val < l2->val)     
         {
             head = l1; 
             l1 = l1->next; 
             
         }
          else                       
         { 
             head = l2; 
             l2 = l2->next;
        }

        ListNode* p = head;     // pointer to form new list
        
        while(l1!=NULL&&l2!=NULL)
        {
            if(l1->val < l2->val)
            {
                p->next = l1;
                l1 = l1 ->next;
            }
            
            else 
            {
                p->next = l2;
                l2 = l2 ->next;
            }
            p = p->next;
        }
        
        if(l1)
        {
            p->next = l1;
        }
        
        else
        {
            p->next = l2;
        }
        
        
        
        return head;
    }
};

 

递归c++解法：

class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        if(l1 == NULL) return l2;
        if(l2 == NULL) return l1;

        if(l1->val < l2->val) {
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        } else {
            l2->next = mergeTwoLists(l2->next, l1);
            return l2;
        }
    }
};

 

python递归解决方式：

def mergeTwoLists(self, l1, l2):
    if not l1:
        return l2
    elif not l2:
        return l1
    else:
        if l1.val <= l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2

python非递归：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param {ListNode} l1
    # @param {ListNode} l2
    # @return {ListNode}
    def mergeTwoLists(self, l1, l2):
        p1 = l1
        p2 = l2
        guard = ListNode(0)
        q = guard
        while p1 is not None and p2 is not None:
            if p1.val <= p2.val:
                q.next = p1
                p1 = p1.next
                q = q.next
            else:
                q.next = p2
                p2 = p2.next
                q = q.next
        if p1 is not None:
            q.next = p1
        if p2 is not None:
            q.next = p2
        return guard.next

 

python递归解决方式2：

If both lists are non-empty, I first make sure a starts smaller, use its head as result, and merge the remainders behind it. Otherwise, i.e., if one or both are empty, I just return what's there.

class Solution:
    def mergeTwoLists(self, a, b):
        if a and b:
            if a.val > b.val:
                a, b = b, a
            a.next = self.mergeTwoLists(a.next, b)
        return a or b