Leetcode 92. Reverse Linked List II

Source: Internet
Author: User

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 .

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

Problem Solving Ideas:

Reverses the variant of the entire list, specifying the start and end points. Because m=1 changes the head node, it joins a dummy head node  1. Find the M-1 node in the original list start: The reversed part will be returned to the node after the change. Move from dummy to m-1  D->1->2->3->4->5->NULL       |      St  2. A partial list of links with a length of L = n-m+1 is reversed starting with P = Start->next.             __________            |                  |            |                 vd->1->2<-3<-4    5->null     & nbsp              |     |           |       St    p          h0 &nbs P        3. Last reply              __________            |                  |    &NBSP       |                 vd->1   2<-3<-4    5->null   &NBSP ;                |________|                

Java code:20160601
/*** Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int X) {val = x;}}*/ Public classSolution { PublicListNode Reversebetween (ListNode head,intMintN) {//iterative//Base Case        if(Head = =NULL|| Head.next = =NULL|| M < 1 | | M >=N) {returnHead; } ListNode Dummy=NewListNode (0); Dummy.next=Head; ListNode Pre=dummy; ListNode cur=Head; //move m-1 to the node before reverse start        intCount = 1;  while(Count <m) {Pre=cur; Cur=Cur.next; Count++; } ListNode beforereverse=Pre; ListNode Startreverse=cur;  while(Count <=N) {ListNode next=Cur.next; Cur.next=Pre; Pre=cur; Cur=Next; Count++; } Beforereverse.next=Pre; Startreverse.next=cur; returnDummy.next; }}


1. http://bangbingsyb.blogspot.com/2014/11/leetcode-reverse-linked-list-ii.html

Leetcode 92. Reverse Linked List II

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