标签:描述 技术分享 lint 情况 public http image param amp
描述
给定一个单链表L: L0→L1→…→Ln-1→Ln,
重新排列后为:L0→Ln→L1→Ln-1→L2→Ln-2→…
必须在不改变节点值的情况下进行原地操作。
样例
思路
先将链表整体一分为二,然后将后半段链表逆序,再依次插入前半段节点中。
/** * Definition of ListNode * class ListNode { * public: * int val; * ListNode *next; * ListNode(int val) { * this->val = val; * this->next = NULL; * } * } */ class Solution { public: /** * @param head: The first node of linked list. * @return: void */ void reorderList(ListNode *head) { // write your code here if(head == NULL || head->next == NULL) { return ; }//if ListNode *p = head, *q = head->next; while(q && q->next) { p = p->next; q = q->next->next; }//while q = p->next; p->next = NULL; //后半段节点先逆序再逐个插入 ListNode *rHead = NULL; while(q) { ListNode *r = q->next; q->next = rHead; rHead = q; q = r; }//while q = rHead; p = head; while(p&&q) { ListNode *rr = q->next; ListNode *lr = p->next; q->next = lr; p->next = q; q = rr; p = lr; }//while } };
标签:描述 技术分享 lint 情况 public http image param amp
原文地址:http://www.cnblogs.com/li1400802003/p/7348531.html