从无头单链表中删除节点

而翻转后，C无法再指向D会出现链表断裂；故在反转前要注意保留D,即C.pNext;

class ListNode
{
 int pValue;
 ListNode pNext;
 
 public ListNode(int pValue,ListNode pNext)
 {
  this.pNext = pNext;
  this.pValue = pValue;
 }
}

public class LinkListEx {
 
 public ListNode reverse(ListNode pHead)
 {
  if (pHead == null || pHead.pNext == null) 
   return pHead;
  ListNode pNodeA = pHead;
  ListNode pNodeB = pHead.pNext;
  ListNode pNodeC = pHead.pNext.pNext;
  
  pHead.pNext = null;
  
  while(pNodeB != null)
  {
   pNodeB.pNext = pNodeA;
   
   pNodeA = pNodeB;
   pNodeB = pNodeC;
   
   if (pNodeB != null)
    pNodeC = pNodeB.pNext;
  }
  
  return pNodeA;
 }
 
 public void delete(ListNode pNode)
 {
  if (pNode == null || pNode.pNext == null) 
   return;
  
  //由于题设中简化了难度，假设pNode既非头结点，也非尾节点，故其实不许进行边界判断
  ListNode pNext    = pNode.pNext;
  
  pNode.pValue = pNext.pValue;
  pNode.pNext  = pNext.pNext;
 }
 
 public static void main(String[] args)
 {
  ListNode p5 = new ListNode(5, null);
  ListNode p4 = new ListNode(4, p5);
  ListNode p3 = new ListNode(3, p4);
  ListNode p2 = new ListNode(2, p3);
  ListNode p1 = new ListNode(1, p2);
  
  LinkListEx linkListEx = new LinkListEx();
//	 linkListEx.print(linkListEx.reverse(p1));
  
  linkListEx.delete(p3);
  linkListEx.print(p1);
 }
 
 public void print(ListNode pHead)
 {
  while(pHead != null)
  {
   System.out.println(pHead.pValue + "==>");
   pHead = pHead.pNext;
  }
 }

}

从无头单链表中删除节点

原文：http://blog.csdn.net/woliuyunyicai/article/details/46483305