链表3

1、合并链表

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

2、解答： 由于两个链表是有序的，所以就比较好解决

3、C++代码

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(l1 == NULL) 
            return l2;
        
        if(l2 == NULL)
            return l1;
        
        ListNode *head = NULL;
        
        if(l1->val < l2->val){
            head = l1;
            l1 = l1->next;
            
        }else{
            head = l2;
            l2 = l2->next;
        }
            
        
        ListNode *p = head;        // 新建个p结点，容易多了
        
        while(l1&&l2){
            if(l1->val < l2->val){
                p->next = l1;
                l1 = l1->next;
            }else{
                p->next = l2;
                l2 = l2->next;
            }
            p = p->next;            //这句不要忘记写了
        }
        
      if(l1){               //l1 不为空的时候
          p->next = l1;
      }else{                //l1 为空
          p->next = l2;
      }
        
     return head;
    }
};

python代码

class Solution:
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """  
        if not l1:
            return l2
        if not l2:
            return l1
        
        if l1.val < l2.val:
            head = l1
            l1 = l1.next
        else:
            head = l2
            l2 = l2.next
        
        p = head
        
        while l1 and l2:
            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

递归解决更简单

    

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(l1, l2->next);
            return l2;
        }