【leetcode】Merge Two Sorted Lists

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.

 

每次找到两个链表中小的一个，把他连接在后面，直到连接完成

 

思路1：利用递归方法解：递归思路：

       

        if(l1->val<=l2->val)

        {

            head=l1;

            head->next=mergeTwoLists(l1->next,l2);

        }

        else

        {

            head=l2;

            head->next=mergeTwoLists(l1,l2->next);

        }

 

 

 

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

 

 

 

非递归解法：

 

/**
 * 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(l1==NULL)
        {
            return l2;
        }
       
        if(l2==NULL)
        {
            return l1;
        }
       
        ListNode *head=new ListNode(0);
        ListNode *pre=head;
       
        while(l1!=NULL||l2!=NULL)
        {
            if(l1==NULL)
            {
                head->next=l2;
                l2=l2->next;
                head=head->next;
                continue;
            }
           
            if(l2==NULL)
            {
                head->next=l1;
                l1=l1->next;
                head=head->next;
                continue;
            }
           
            if(l1->val<=l2->val)
            {
                head->next=l1;
                l1=l1->next;
                head=head->next;
            }
            else
            {
                head->next=l2;
                l2=l2->next;
                head=head->next;
            }
        }
       
        ListNode *ret=pre->next;
        delete pre;
        return pre->next;
       
    }
};