Reverse Linked List | & ||

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。

 /**
  * Definition for ListNode.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int val) {
  *         this.val = val;
  *         this.next = null;
  *     }
  * }
  */
 public class Solution {
     /**
      * @param head: The head of linked list.
      * @return: The new head of reversed linked list.
      */
     public ListNode reverse(ListNode head) {
         if (head == null) return head;
         ListNode prev = null;
         ListNode cur = head;
         ListNode next = null;
         while (cur != null) {
             next = cur.next;
             cur.next = prev;
             prev = cur;
             cur = next;
         }
         return prev;
     }
 }

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。

 /**
  * Definition for ListNode
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */
 public class Solution {
     /**
      * @param ListNode head is the head of the linked list
      * @oaram m and n
      * @return: The head of the reversed ListNode
      */
     public ListNode reverseBetween(ListNode head, int m , int n) {
         if (m == n) return head;
         ListNode tempHead = new ListNode();
         tempHead.next = head;

         // reach the end of the first part
         ListNode partEnd = tempHead;
         for(int k = ; k < m; k++) {
             partEnd = partEnd.next;
         }
         // save it later to connect to the last part.
         ListNode secondPartTail = partEnd.next; 

         // reverse the middle part
         ListNode prev = null;
         ListNode current = partEnd.next;
         ListNode next = null;

         for (int p = ; p <= n - m + ; p++) {
             next = current.next;
             current.next = prev;
             prev = current;
             current = next;
         }
         // connect three parts
         partEnd.next = prev;
         secondPartTail.next = current;

         return tempHead.next;
     }
 }

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