Reverse Linked List | & ||

Posted 2020-08-02 北叶青藤

Reverse Linked List I

Reverse a linked list.

Example

For linked list 1->2->3, the reversed linked list is 3->2->1

分析：

典型的3 pointers 问题。

 1 /**
 2  * Definition for ListNode.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int val) {
 7  *         this.val = val;
 8  *         this.next = null;
 9  *     }
10  * }
11  */ 
12 public class Solution {
17     public ListNode reverse(ListNode head) {
18         if (head == null) return head;
19         ListNode prev = null;
20         ListNode cur = head;
21         ListNode next = null;
22         while (cur != null) {
23             next = cur.next;
24             cur.next = prev;
25             prev = cur;
26             cur = next;
27         }
28         return prev;
29     }
30 }

Reverse Linked List II

Reverse a linked list from position m to n.

Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list.

Example

Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL.

分析：

我们需要把list分成三段。首先得到第一段的最后一个node，然后reverse中间部分。最后把三个部分链接起来。

 1 class Solution {
 2     public ListNode reverseBetween(ListNode head, int m, int n) {
 3         if (m == n) return head;
 4         ListNode tempHead = new ListNode(1);
 5         tempHead.next = head;
 6 
 7         // reach the end of the first part
 8         ListNode firstPartEnd = tempHead;
 9         for (int k = 1; k < m; k++) {
10             firstPartEnd = firstPartEnd.next;
11         }
12         // save it later to connect to the last part.
13         ListNode secondPartEnd = firstPartEnd.next;
14 
15         // reverse the middle part
16         ListNode prev = null;
17         ListNode curr = firstPartEnd.next;
18         ListNode next = null;
19 
20         for (int p = 1; p <= n - m + 1; p++) {
21             next = curr.next;
22             curr.next = prev;
23             prev = curr;
24             curr = next;
25         }
26         // connect three parts
27         firstPartEnd.next = prev;
28         secondPartEnd.next = current;
29 
30         return tempHead.next;
31     }
32 }

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