Leetcode: 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.

比较简单。

/**
 * 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;
       if (l1->val <= l2->val) {
           head = l1;
           l1 = l1->next;
       }
       else {
           head = l2;
           l2 = l2->next;
       }
       ListNode *cur = head;
       while (l1 != NULL && l2 != NULL) {
           if (l1->val <= l2->val) {
               cur->next = l1;
               l1 = l1->next;
           }
           else {
               cur->next = l2;
               l2 = l2->next;
           }
           cur = cur->next;
       }
       
       if (l1 != NULL) {
           cur->next = l1;
       }
       if (l2 != NULL) {
           cur->next = l2;
       }
       
       return head;
    }
};

======================第二次========================

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