Given 1->2->3->4->5->NULL
,
return 1->3->5->2->4->NULL
.
就是将序号为单数的放在前面,而序号为偶数的放在后面
我的方法是讲序号为偶数的的插入到链表末尾。
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* oddEvenList(ListNode* head) { 12 if(!head) return head;//空链表返回 13 ListNode* last = head; 14 int cnt = 1; 15 for(; last->next; last = last->next, ++cnt); 16 if(cnt < 3) return head;//链表长度小于3返回 17 ListNode* now = head; 18 ListNode* end = last; 19 20 for(int i = 0; i< cnt/2; now = now->next, ++i){ 21 22 ListNode* next = now->next; 23 now->next = next->next; 24 25 end->next = next; 26 next->next = NULL; 27 28 end = next; 29 30 } 31 return head; 32 } 33 };