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;
}