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.

递归算法：

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

非递归算法：

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;}