描述:
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->NULL
, m = 2 and n =
4,
return 1->4->3->2->5->NULL
.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
思路:
这种题目,举个例子能让思路更加清晰,通过在草纸上演算可知,题目要分两种情况,m==1和m>1的情况,然后就是围绕这两种情况展开讨论,删除后面的结点,然后将后面的结点添加到前面,一次搞定,bravo!
代码:
public ListNode reverseBetween(ListNode head, int m, int n) { if (head==null) { return null; } ListNode p =head,q=head,temp=null; int i=0; for(i=0;i<m-1;i++) q=q.next; for(i=0;i<m-2;i++) p=p.next; if(m==1) { for(i=0;i<n-m;i++) { //delete the node in the list temp=q.next; q.next=temp.next; //insert the node in the list temp.next=p; p=temp; } head=p; }else { for(i=0;i<n-m;i++) { //delete the node in the list temp=q.next; q.next=temp.next; //insert the node in the list temp.next=p.next; p.next=temp; } } return head; }
结果:
时间: 2024-11-05 13:32:06