328. Odd Even Linked List

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if(NULL == head)
            return NULL;
        ListNode *odd = head,*evenhead = head->next,*even = evenhead;
        while(even && even->next)
        {
            odd->next = odd->next->next;
            even->next = even->next->next;
            
            odd = odd->next;
            even = even->next;
        }
        
        odd->next = evenhead;
        
        return head;
    }
};

链表的题相对简单，但有时候会卡壳。难点就在于遍历整个链表时终止条件。