题目原型:
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.
基本思路:
这个题目较上一个题的不同之处在于,本题中要删除重复元素,采用的思路是先找到head指针,就是头指针,例如:1->1->1->2->3 头指针就是指向2的指针,1->2->3->3->4->4->5头指针就是指向1的指针,以此类推,然后从
头指针开始扫描,类似上一题。
public ListNode deleteDuplicates(ListNode head) {
if(head==null)
return null;
ListNode p = head,pre;
ListNode nextNode = p.next;
boolean hasFound = false;
int temp = 0;
int flag = 0;
if(nextNode==null)
return head;
while(nextNode!=null)
{
if(p.val!=nextNode.val&&flag==0)
{
head = p;
hasFound = true;
break;
}
else if(p.val!=nextNode.val&&flag==1)
flag = 0;
else if(p.val==nextNode.val)
flag = 1;
p = nextNode;
nextNode = p.next;
}
if(hasFound==false)
{
if(p!=null&&flag==0)
head = p;
else
return null;
}
p = head.next;
pre = head;
if(p!=null)
temp = pre.val;
while(p!=null)
{
if(temp!=p.val)
{
if(p.next!=null&&p.next.val!=p.val)
{
pre.next = p;
pre = p;
}
if(p.next==null)
pre.next = p;
temp = p.val;
flag = 0;
}
else
{
flag = 1;
}
p = p.next;
}
if(flag==1)
pre.next = null;
return head;
}
Remove Duplicates from Sorted List II
原文:http://blog.csdn.net/cow__sky/article/details/19689335