【Leetcode】【Easy】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.

解题：

普通做法三步：

1、常规边界判定

2、由于最终要返回合成的链表头部，因此需要先比较l1和l2首个结点值大小，确定ret值；

3、新建一个cur指针操作要返回的链表，串起l1和l2；

记录：

1、指针赋值的关注点在等号的右边。指针名是变量名，随着赋值操作随时改变，不会产生连带影响；

1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
12         ListNode *ret = NULL;
13         ListNode *cur = NULL;
14         
15         if (!l1) return l2;
16         if (!l2) return l1;
17             
18         if (l1->val <= l2->val) {
19             ret = l1;
20             l1 = l1->next;
21         } else { 
22             ret = l2;
23             l2 = l2->next;
24         }
25         
26         cur = ret;
27         while (l1 && l2) {
28             if (l1->val <= l2->val) {
29                 cur->next = l1;
30                 l1 = l1->next;
31             } else {
32                 cur->next = l2;
33                 l2 = l2->next;
34             }
35             cur = cur->next;
36         }
37         
38         if (!l1) 
39             cur->next = l2;
40         else 
41             cur->next = l1;
42         
43         return ret;
44     }
45 };      

文艺做法：

使用二重指针解决了判定链表头大小的步骤

类似另一道简单题Remove Nth Node From End of List

Linus once said " People who understand pointers just use a“pointer to the entry pointer”"

1 class Solution {
 2 public:
 3     ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
 4         ListNode* head = NULL, **t = &head;
 5         while(l1 != NULL && l2 != NULL) {
 6             if(l1->val < l2->val) {
 7                 *t = l1;
 8                 l1 = l1->next;
 9             } else {
10                 *t = l2;
11                 l2 = l2->next;
12             }
13             t = &((*t)->next);
14         }
15         *t = (l1 != NULL ? l1 : l2);
16         return head;
17     }
18 };