标签：des style class blog code ext
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You must do this in-place without altering the nodes‘ values.
For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.
思路：用队列或者栈将整个链表存储起来，并计算链表长度。然后将队尾或者栈顶元素取出插入到对头元素的后面。这样就可以完成要求了。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: void reorderList(ListNode *head) { deque<ListNode *> pList; ListNode *pNode=head; ListNode *pCur=head; int count=0; while(pNode!=NULL) { pList.push_back(pNode); pNode=pNode->next; count++; } for(int i=0;i<count/2;i++) { ListNode *temp=pList.back(); pList.pop_back(); temp->next=pCur->next; pCur->next=temp; pCur=temp->next; } if(head!=NULL) pCur->next=NULL; } };
标签：des style class blog code ext
原文地址：http://www.cnblogs.com/awy-blog/p/3807811.html
