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ReverseLinkedList：
public class ReverseLinkedList
{
public ListNode reverseList(ListNode head)
{
if(head == null || head.next == null)
{
return head;
}
ListNode pPre = new ListNode(0);
pPre.next = head;
ListNode curr = head;
ListNode pNext = null;
while(curr != null)
{
pNext = curr.next;
curr.next = pPre;
pPre = curr;
curr = pNext;
}
head.next = null;
return pPre;
}
}
ReverseLinkedList2:
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.
算法分析：其实就是反转链表加了个条件，先找到第m-1个元素，m个元素，第n+1个元素，反转后再连接起来。
public class ReverseLinkedList2
{
public ListNode reverseBetween(ListNode head, int m, int n)
{
ListNode temp1 = new ListNode(0);
//新的头结点
ListNode newHead = temp1;
temp1.next = head;
//找到第m-1个节点
for(int i = 1; i < m; i ++)
{
temp1 = temp1.next;
}
ListNode temp2 = new ListNode(0);
temp2.next = head;
//找到第n+1个节点
for(int i = 0; i < n+1; i ++)
{
temp2 = temp2.next;
}
//第m个节点
ListNode temp = temp1.next;
//反转第m到n节点
ListNode pPre = new ListNode(0);
pPre.next = temp;
ListNode pCurr = temp1.next;
ListNode pNext = null;
while(pCurr != temp2)
{
pNext = pCurr.next;
pCurr.next = pPre;
pPre = pCurr;
pCurr = pNext;
}
temp1.next = pPre;
temp.next = temp2;
return newHead.next;
}
}
ReverseLinkedList，ReverseLinkedList2，链表反转问题
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原文地址：http://www.cnblogs.com/masterlibin/p/5824156.html
