在O(1)时间内删除链表结点 【微软面试100题 第六十题】
题目要求:
给定链表的头指针和一个结点指针,在O(1)时间删除该结点。
参考资料:剑指offer第13题。
题目分析:
有几种情况:
1.删除的结点是头结点,且链表不止一个结点;
2.删除的结点是头结点,且链表只有一个结点;
3.删除的结点是尾结点,且链表不止一个结点;
4.删除的结点不是头也不是尾结点;
对于第四种情况(普遍情况),见如下图分析:
对于第三种情况,时间复杂度为O(n).其他情况时间复杂度为O(1).则总的平均复杂度为[(n-1)*O(1)+O(n)]/n = O(1).
代码实现:
#include <iostream> #include <stack> using namespace std; typedef struct ListNode { struct ListNode *next; int data; }ListNode; void InitList(ListNode **head1,ListNode **toBeDelete); void DeleteListNode(ListNode **pHead,ListNode *pToBeDelete); void PrintList(ListNode *list); int main(void) { ListNode *h,*toBeDelete; InitList(&h,&toBeDelete); PrintList(h); cout <<"删除5" << endl; DeleteListNode(&h,toBeDelete); PrintList(h); return 0; } void PrintList(ListNode *list) { while(list!=NULL) { cout << list->data << "->"; list = list->next; } cout << "NULL"; cout <<endl; } void DeleteListNode(ListNode **pHead,ListNode *pToBeDelete) { if(!pHead || !(*pHead) || !pToBeDelete) return ; //情况1 if(*pHead==pToBeDelete && pToBeDelete->next != NULL) { *pHead = pToBeDelete->next; delete pToBeDelete; } //情况4 else if(pToBeDelete->next != NULL) { ListNode *pNext = pToBeDelete->next; pToBeDelete->data = pNext->data; pToBeDelete->next = pNext->next; delete pNext; pNext = NULL; } //情况2 else if(*pHead==pToBeDelete) { delete pToBeDelete; pToBeDelete = NULL; *pHead = NULL; } //情况3 else { ListNode *pNode = *pHead; while(pNode->next != pToBeDelete) pNode = pNode->next; pNode->next = NULL; delete pToBeDelete; pToBeDelete = NULL; } } //head:1-->5-->9-->NULL void InitList(ListNode **head1,ListNode **toBeDelete) { ListNode *tmp = new ListNode; tmp->data = 1; *head1 = tmp; tmp = new ListNode; tmp->data = 5; (*head1)->next = tmp; *toBeDelete = tmp; ListNode *tmp1 = new ListNode; tmp1->data = 9; tmp1->next = NULL; tmp->next = tmp1; }
很多时候不是我们做不好,而是没有竭尽全力......
posted on 2014-11-16 11:49 tractorman 阅读(328) 评论(0) 编辑 收藏 举报