Reverse Linked List II

来源：互联网发布：淘宝网家居拖鞋编辑：程序博客网时间：2024/05/20 17:25

<span style="font-size:14px;">这一道题，最好先做之前的  Reverse Linked ListI</span>

之后把他分成几个节点

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */ int LengthList(struct ListNode* head) {if(head==NULL)return 0;int length=0;struct ListNode* pCurrent=head;while(pCurrent!=NULL){length++;pCurrent=pCurrent->next;}//free(pCurrent);return length;} struct ListNode* reverseList(struct ListNode* head) {struct ListNode* newhead;newhead=NULL;//这样就不需要翻转了if (head==NULL||head->next==NULL){return head;}struct ListNode* pCurrent=head;  //一样的点while(pCurrent!=NULL){struct ListNode* tmp=pCurrent;pCurrent=pCurrent->next;tmp->next=newhead;   //把节点放在节点前面,插入到链表的最前端newhead=tmp;}return newhead;}struct ListNode* reverseBetween(struct ListNode* head, int m, int n) {int length;struct ListNode* pCurrent=head;length=LengthList(pCurrent);int diff=n-m;if (diff==0){return head;}if (m==1){if (length==n){head=reverseList(head);return head;}if (n<length){pCurrent=head;while (diff--)   //{pCurrent=pCurrent->next;}struct ListNode* pCurrentNhead;pCurrentNhead=pCurrent->next;struct ListNode* pCurrentMTail=pCurrent;pCurrent->next=NULL; //first tail node.head=reverseList(head);pCurrent=head;while(pCurrent->next!=NULL)  //走到反转的地方n头结点的前面..  //另外一种方法m都头结点已经算是尾节点了。{pCurrent=pCurrent->next;}pCurrent->next=pCurrentNhead;return head; }}else{if (length==n)     {struct ListNode* pCurrentHeadTail=head;pCurrent=head;     m--;    while(m--)   {   pCurrentHeadTail=pCurrent;   pCurrent=pCurrent->next;   }   //分段   pCurrentHeadTail->next=NULL;   struct ListNode* pCurrentHeadm=pCurrent;    pCurrent=pCurrentHeadm;   pCurrentHeadm=reverseList(pCurrentHeadm);   pCurrentHeadTail->next=pCurrentHeadm;   return head;  }if (n<length){struct ListNode* pCurrentHeadTail=head;pCurrent=head;m--;while(m--){pCurrentHeadTail=pCurrent;pCurrent=pCurrent->next;}//分段pCurrentHeadTail->next=NULL;struct ListNode* pCurrentHeadm=pCurrent; while (diff--){pCurrent=pCurrent->next;}struct ListNode* pCurrentNhead;pCurrentNhead=pCurrent->next;struct ListNode* pCurrentMTail=pCurrent;pCurrent->next=NULL;pCurrentMTail=pCurrentHeadm;pCurrentHeadm=reverseList(pCurrentHeadm);pCurrentHeadTail->next=pCurrentHeadm;pCurrentMTail->next=pCurrentNhead;return head;}}}

0 0