Reorder List--简单粗暴的水题

题目大意

好吧，这题之前看错题目了，以为是按照奇偶依次提取然后将偶序列倒序后插入奇序列，那写得我叫一个纠结，后来仔细读了一遍题目才发现不是酱紫。。题目大意是将一个链表的后半段倒序后依次插入前半段的空格中，简单粗暴。

思路

显然解法也是简单粗暴，顺序就是：截取后半段->倒序->插入前半段，详见代码

总结

看题得仔细，另外插入时注意对空结点的处理

代码

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {    public:        ListNode *reverse(ListNode *cur, ListNode *fa)//翻转链表         {            if (cur == NULL)return fa;                ListNode *temp = reverse(cur->next, cur);            cur->next = fa;            return temp;        }        void reorderList(ListNode *head)        {            if (head == NULL)return ;            ListNode *headE, *headO, *f_p, *s_p;            f_p = s_p = headE = headO = head;//快慢指针             int count = 0;            while (f_p->next != NULL){                if (count % 2)s_p = s_p -> next;                f_p = f_p -> next;                count ++;            }            headO = s_p->next;//后半段序列的头             s_p->next = NULL;//断链             headO = reverse(headO, NULL);//翻转             ListNode *odd = headO;            ListNode *even = headE;            while (odd != NULL || even != NULL){//插入                 ListNode *tmp = even;                if (even != NULL) even = even->next;                if (odd != NULL)  tmp->next = odd;                   tmp = odd;                if (odd != NULL)  odd = odd-> next;                if (even != NULL) tmp->next = even;                     }                    }        };