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.

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
一：递归写法
，感觉比较难理解==

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;    }}

};

二：非递归写法

ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {    if(NULL == l1) return l2;    if(NULL == l2) return l1;    ListNode* head=NULL;    // head of the list to return    // find first element (can use dummy node to put this part inside of the loop)    if(l1->val < l2->val)       { head = l1; l1 = l1->next; }    else                        { head = l2; l2 = l2->next; }    ListNode* p = head;     // pointer to form new list    // I use && to remove extra IF from the loop    while(l1 && l2){        if(l1->val < l2->val)   { p->next = l1; l1 = l1->next; }        else                    { p->next = l2; l2 = l2->next; }        p=p->next;    }    // add the rest of the tail, done!    if(l1)  p->next=l1;    else    p->next=l2;    return head;}