[Leetcode][JAVA] Merge Two Sorted Lists & Sort List

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.

两个排好序的链表拼接，只要用两个指针遍历链表即可. 可以借助一个helper指针来进行开头的合并。如果有一个遍历完，则直接把另一个链表指针之后的部分全部接在后面：

 1 public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
 2         ListNode helper = new ListNode(0);
 3         ListNode cur1 = l1;
 4         ListNode cur2 = l2;
 5         ListNode cur = helper;
 6         while(cur1!=null || cur2!=null)
 7         {
 8             if(cur1==null)
 9             {
10                 cur.next = cur2;
11                 break;
12             }
13             if(cur2==null)
14             {
15                 cur.next = cur1;
16                 break;
17             }
18             if(cur1.val<=cur2.val)
19             {
20                 cur.next = cur1;
21                 cur1=cur1.next;
22             }
23             else
24             {
25                 cur.next = cur2;
26                 cur2=cur2.next;
27             }
28             cur=cur.next;
29         }
30         return helper.next;
31     }

Sort List:

Sort a linked list in O(n log n) time using constant space complexity.

有了Merged Two Sorted Lists的基础，这题就很好写了。提到O(nlogn)的排序算法，马上想到快排和合并排序，不过快排不稳定而且对于链表来说交换不方便，所以还是决定采用merge sort.

对于链表，merge sort中的split过程可以使用快慢指针实现，比较简单易懂：

 1 public ListNode sortList(ListNode head) {
 2         return mergeSort(head);
 3     }
 4
 5     public ListNode mergeSort(ListNode start)
 6     {
 7         if(start==null) return null;
 8         if(start.next==null) return start;
 9
10         ListNode slow = start;
11         ListNode fast = start;
12         while(fast!=null && fast.next!=null && fast.next.next!=null)
13         {
14             slow = slow.next;
15             fast = fast.next.next;
16         }
17         ListNode second = slow.next;
18         slow.next = null;
19         return mergeTwoLists(mergeSort(start), mergeSort(second));
20     }
21     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
22         ListNode helper = new ListNode(0);
23         ListNode cur1 = l1;
24         ListNode cur2 = l2;
25         ListNode cur = helper;
26         while(cur1!=null || cur2!=null)
27         {
28             if(cur1==null)
29             {
30                 cur.next = cur2;
31                 break;
32             }
33             if(cur2==null)
34             {
35                 cur.next = cur1;
36                 break;
37             }
38             if(cur1.val<=cur2.val)
39             {
40                 cur.next = cur1;
41                 cur1=cur1.next;
42             }
43             else
44             {
45                 cur.next = cur2;
46                 cur2=cur2.next;
47             }
48             cur=cur.next;
49         }
50         return helper.next;
51     }