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.

思路.首先判断是否有空链表，如果有，则直接返回另一个链表，如果没有，则开始比较两个链表的当前节点，返回较小的元素作为前驱，并且指针向后移动一位，再进行比较，如此循环，知道一个链表的next指向NULL，将另一个链表的后序元素进行连接即可。

迭代：

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

循环：

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