剑指Offer16合并两个排序的链表

Posted 2020-11-27 blog-cpc

题目描述

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

解法一：使用ArrayList

 1 public static ListNode Merge(ListNode list1,ListNode list2) {
 2         if(list1==null||list2==null){
 3             if(list1==null){
 4                 return list2;
 5             }else{
 6                 return list1;
 7             }
 8         }
 9         ArrayList<ListNode> res = new ArrayList<>();
10         while (list1!=null&&list2!=null){
11             if(list1.val<=list2.val){
12                 res.add(new ListNode(list1.val));
13                 list1=list1.next;
14             }else{
15                 res.add(new ListNode(list2.val));
16                 list2=list2.next;
17             }
18         }
19         while(list1!=null){
20             res.add(new ListNode(list1.val));
21             list1=list1.next;
22         }
23         while (list2!=null){
24             res.add(new ListNode(list2.val));
25             list2=list2.next;
26         }
27         //建立节点间的next联系
28         for(int i = 0;i<res.size()-1;i++){
29             res.get(i).next = res.get(i+1);
30         }
31         return res.get(0);
32     }

解法二：链表

 1 public static ListNode Merge01(ListNode list1,ListNode list2) {
 2         ListNode newHead = new ListNode(-1);
 3         ListNode current = newHead;
 4         while (list1 != null && list2 != null) {
 5             if (list1.val < list2.val) {
 6                 current.next = list1;
 7                 list1 = list1.next;
 8             } else {
 9                 current.next = list2;
10                 list2 = list2.next;
11             }
12             current = current.next;
13         }
14         if (list1 != null){
15             current.next = list1;
16         }
17         if (list2 != null) {
18             current.next = list2;
19         }
20         return newHead.next;
21     }

解法三：递归

 1 public ListNode Merge02(ListNode list1,ListNode list2) {
 2         if (list1 == null) {
 3             return list2;
 4         }
 5         if (list2 == null) {
 6             return list1;
 7         }
 8         if (list1.val < list2.val) {
 9             list1.next = Merge(list1.next, list2);
10             return list1;
11         } else {
12             list2.next = Merge(list1, list2.next);
13             return list2;
14         }
15     }

初始化链表：

 1 public static class ListNode {
 2         int val;
 3         ListNode next = null;
 4         ListNode(int val) {
 5             this.val = val;
 6         }
 7     }
 8 public static ListNode createList(int[] list){
 9         ListNode head = new ListNode(-1);
10         ListNode current=head;
11         for(int i=0;i<list.length;i++){
12             current.next=new ListNode(list[i]);
13             current=current.next;
14         }
15         return head.next;
16     }

测试：

 1 public static  void main(String[] args) {
 2         int[] list1={1,3,5};
 3         int[] list2={2,4,6};
 4         ListNode root1 = createList(list1);
 5         ListNode root2 = createList(list2);
 6         ListNode merge = Merge(root1, root2);
 7         while (merge!=null){
 8             System.out.print(merge.val+" ");
 9             merge=merge.next;
10         }
11     }
12 输出：1 2 3 4 5 6