本博文内容为单链表相关的笔试题,转载请注明转载处,否则必究
1. 合并两个有序的单链表成一个有序的单链表
方法分为递归实现与非递归实现,两种方法都不额外开辟 内存空间
链表的数据结构在本博客的单链表逆转,约瑟夫环等
递归实现:
//递归实现合并两个有序单链表LinkNode* merge_list(LinkNode *pHead1, LinkNode *pHead2){if(pHead1==NULL)return pHead2;if(pHead2==NULL)return pHead1;if(pHead1==NULL && pHead2==NULL)return NULL;LinkNode *pMergedHead=NULL;if(pHead1->value<pHead2->value){pMergedHead=pHead1;pMergedHead->next = merge_list(pHead1->next, pHead2);}else {pMergedHead=pHead2;pMergedHead->next=merge_list(pHead1, pHead2->next);}return pMergedHead;}
非递归实现:
//非递归实现合并两个有序单链表(不额外开辟空间)LinkNode* non_merge_list(LinkNode *pHead1, LinkNode *pHead2){if(pHead1==NULL)return pHead2;if(pHead2==NULL)return pHead1;if(pHead1==NULL && pHead2==NULL)return NULL;LinkNode *pMergedHead = NULL;LinkNode *q=NULL;if(pHead1->value<pHead2->value){pMergedHead=pHead1;pHead1=pHead1->next;}else{pMergedHead=pHead2;pHead2=pHead2->next;}q=pMergedHead;while(pHead1 && pHead2){if(pHead1->value<pHead2->value){q->next=pHead1;pHead1=pHead1->next;}else{q->next=pHead2;pHead2=pHead2->next;}q=q->next;}if(pHead1){while(pHead1){q->next=pHead1;q=q->next;pHead1=pHead1->next;}}if(pHead2){while(pHead2){q->next=pHead2;q=q->next;pHead2=pHead2->next;}}return pMergedHead;}
2 输出单链表中的中间元素(若链表节点个数为偶数,则输出中间两个的任意一个)
思路:利用两个指针从头节点开始遍历,一个走一步,一个走两步,当一次走两步的指针走到链表末尾时,此时一次走一步的指针就指向链表的中间节点
代码如下:
LinkNode* print_mid_node(LinkNode *pHead){LinkNode *pOne = pHead, *pTwo = pHead;while(1){pOne = pOne->next;pTwo = pTwo->next->next;if(pTwo==NULL || pTwo->next==NULL)return pOne;}}
3 判断单恋表是否有环
思路与第二题一样,只是结束条件不一样,如果当一次走一步的指针等于一次走两步的指针时,则表示该链表有环
代码如下:
bool is_circle_list(LinkNode *pHead){LinkNode *pOne = pHead, *pTwo = pHead;while(1){pOne = pOne->next;pTwo = pTwo->next->next;if(pOne == pTwo)return true;if(pTwo==NULL || pTwo->next==NULL)return false;}}