在O(1)时间内删除链表结点删除链表中重复节点

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题意描述：给定单向链表的头指针和一个结点指针，定义一个函数在O(1)时间删除该结点

解题思路：要求在O(1)时间内删除结点，则可以这样考虑：将给定结点后继结点的值赋给给定结点，然后删除给定结点的后继结点即可

void DeleteNode(ListNode** pHead, ListNode* pToBeDeleted) {if (!pHead || !pToBeDeleted)return;if (pToBeDeleted->m_pNext != NULL) {//如果要删除的结点不是尾结点ListNode* pNode = pToBeDeleted->m_pNext;pToBeDeleted->m_nValue = pNode->m_nValue;pToBeDeleted->m_pNext = pNode->m_pNext;delete pNode;pNode = NULL;}else if(*pHead == pToBeDeleted){//如果要删除的结点只有一个头结点delete pToBeDeleted;pToBeDeleted = NULL;pHead = NULL;}else {//链表有多个结点，且删除的是尾结点ListNode* p = *pHead;while (p->m_pNext != pToBeDeleted)p = p->m_pNext;p->m_pNext = NULL;delete pToBeDeleted;pToBeDeleted = NULL;}}

题意描述：删除链表中重复节点

1、重复节点只保留一个

/*public class ListNode {int val;ListNode next = null;ListNode(int val) {this.val = val;}
}
*/
public class Solution {public ListNode deleteDuplication(ListNode pHead){if (pHead == null || pHead.next == null) {return pHead;}ListNode newHead = pHead;ListNode curNode = pHead.next;while (curNode != null) {if(curNode.val == pHead.val) {curNode = curNode.next;pHead.next = curNode;} else {pHead = pHead.next;curNode = curNode.next;}}return newHead;}
}

2、重复节点不保留

/*public class ListNode {int val;ListNode next = null;ListNode(int val) {this.val = val;}
}
*/
public class Solution {public ListNode deleteDuplication(ListNode pHead){if (pHead == null || pHead.next == null) {return pHead;}ListNode newHead = new ListNode(-1);newHead.next = pHead;ListNode last = newHead;ListNode cur = pHead;while(cur != null && cur.next != null) {if(cur.val == cur.next.val) {int val = cur.val;while (cur != null && cur.val == val) {cur = cur.next;}last.next = cur;} else {last = cur;cur = cur.next;}}return newHead.next;}
}

 

 

 

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