https://leetcode.com/problems/merge-two-sorted-lists/description/
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.
Example:
Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4
1 // time o(n): while loop sapce o(1): one stack
2 public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
3 ListNode dummy = new ListNode(0);
4 ListNode curr = dummy;
5 // l1 l2 two heads w e dont need to keep, so we use them as counter: so we dont need to check l1.next or l2.next
6 while (l1!=null && l2 != null){
7 if (l1.val <= l2.val){
8 curr.next = l1 ;
9 l1 = l1.next ;
10 }else{
11 curr.next = l2;
12 l2 = l2.next ;
13 }
14 curr = curr.next ;
15 }
16 //either side is null, check the other
17 if (l1 != null){
18 curr.next = l1 ;
19 }
20 if (l2 != null){
21 curr.next = l2 ;
22 }
23 //serve as default for if either l1 is null or l2 is null or reach the end
24 return dummy.next;
25 }
26
27 /*
28 * Input: 1->2->4, 1->3->4
29 Output: 1->1->2->3->4->4
30 * */
31 // time: o(n) space: o(n)-recursive on n stacks
32 public ListNode mergeTwoLists2(ListNode l1, ListNode l2) {
33 /* recursive: subproblem: l1,l2 -> l1.next or l2.next but not l1.next && l2.next
34 * rule: l1.val <= l2 l1.next->l1.next
35 * l2.val< l1 l2.next->l2.next
36 * base l1 == null return l2; l2 == null return l1
37 * */
38 if (l1 == null) return l2 ;
39 if (l2 == null) return l1 ;
40 if (l1.val <= l2.val){
41 l1.next = mergeTwoLists2(l1.next, l2);
42 return l1 ;
43 } else{
44 l2.next = mergeTwoLists2(l1, l2.next) ;
45 return l2 ;
46 }
47 //这样写的错误在于，L1 L2 只链接了一个点
48 // ListNode newHead = mergeTwoLists2(l1.next, l2.next) ;
49 // if (l1.val <= l2.val) {
50 // l1.next = newHead ;
51 // return l1 ;
52 // } else{
53 // l2.next = newHead ;
54 // return l2 ;
55 // }
56 }
