Leetcode 21. Merge Two Sorted Lists（easy）

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.

 递归实现： 

/**
 * 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(l1, l2 -> next);
         return l2;
     }
        
    }
};

 非递归实现：

 

/**
 * 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(0);   // 新建一个结点，不能是指向空的指针，否则它的指向不会改变
        ListNode * cur = &head;
        while (l1 && l2){
            if (l1 -> val <= l2 -> val){
                cur -> next = l1;
                l1 = l1 -> next;
            }else{
                cur -> next = l2;
                l2 = l2 -> next;
            }
            cur = cur -> next;
        }
        if (l1){
            cur -> next = l1;
        }
        if (l2){
            cur -> next = l2;
        }
        return head.next;  // 注意head此时为一个实体对象，不是一个指针，所以要用.来指示next.为什是next??
    }
};