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删除链表中重复的结点
题目描述:
在一个排序的链表中,存在重复的结点,请删除该链表中重复的结点,重复的结点不保留,返回链表头指针。 例如,链表1->2->3->3->4->4->5 处理后为 1->2->5
###解题思路:
解法一:递归
public class Solution{
public ListNode deleteDuplication(ListNode pHead){
if (pHead == null || pHead.next == null )
return pHead;
//处理重复结点
if (pHead.val == pHead.next.val){
ListNode cur = pHead.next;
while (cur != null && cur.val == pHead.val){
cur = cur.next;
}
return deleteDuplication(cur);
}
//该结点不重复,递归到下一个结点
pHead.next = deleteDuplication(pHead.next);
return pHead;
}
}
解法二:非递归 三指针
public class Solution {
public ListNode deleteDuplication(ListNode pHead){
//设置一个头结点,让从一开始就节点重复更好操作
ListNode first = new ListNode(-1);
first.next = pHead;
ListNode p = pHead;
ListNode last = first;//用户删除重复节点的节点
while (p != null && p.next != null) {
if (p.val == p.next.val) {
int val = p.val;
while (p!= null&&p.val == val){
p = p.next;
}
last.next = p;//删除重复节点
} else {
last = p;
p = p.next;
}
}
return first.next;
}
}