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  • [Leetcode] Reverse Linked List II

    Reverse a linked list from position m to n. Do it in-place and in one-pass.

    For example:
    Given 1->2->3->4->5->NULLm = 2 and n = 4,

    return 1->4->3->2->5->NULL.

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

    Solution:

     1 /**
     2  * Definition for singly-linked list.
     3  * public class ListNode {
     4  *     int val;
     5  *     ListNode next;
     6  *     ListNode(int x) {
     7  *         val = x;
     8  *         next = null;
     9  *     }
    10  * }
    11  */
    12 public class Solution {
    13     public ListNode reverseBetween(ListNode head, int m, int n) {
    14         if(m==n)
    15             return head;
    16         ListNode dummy=new ListNode(-1);
    17         dummy.next=head;
    18         ListNode pointer=dummy;
    19         int cnt=m-1;
    20         while(cnt!=0){
    21             pointer=pointer.next;
    22             cnt--;
    23         }
    24         ListNode l1,l2,l3;
    25         l1=pointer.next;
    26         ListNode last=l1;
    27         l2=l1.next;
    28         l3=l2.next;
    29         cnt=n-m-1;
    30         while(cnt!=0){
    31             l2.next=l1;
    32             
    33             l1=l2;
    34             l2=l3;
    35             l3=l3.next;
    36             cnt--;
    37         }
    38         l2.next=l1;
    39         last.next=l3;
    40         pointer.next=l2;
    41         return dummy.next;
    42     }
    43 }
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  • 原文地址:https://www.cnblogs.com/Phoebe815/p/4101320.html
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