LeetCode 328. Odd Even Linked List

一开始理解错题意，以为是根据节点内的值是奇偶来分成不同group，实际上是节点所在位置的奇偶（位置从1开始算起）。

 

思路正确后仍然RE的代码：

class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if(head == NULL) return head;
        if(head->next == NULL) return head;
        
        ListNode* odd = head, * even = head->next;
        odd->next = NULL;
        even->next = NULL;
        ListNode* o_head = odd, * e_head = even;
        while(odd && even && even->next){
            odd->next = even->next;
            odd = odd->next;
            
            even->next = odd->next;
            even = even->next;
        }
        even->next = NULL;
        odd->next = e_head->next;
        return o_head->next;
    }
};

起因是8、9行，因为odd和even本身已经是head和head->next两个节点，再设置next为NULL就相当于截断了原链表，even->next=NULL，进不了while循环。如果把18、19注释掉，不会RE。

 

class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if(head == NULL) return head;
        if(head->next == NULL) return head;
        
        ListNode* odd = head, * even = head->next;
        ListNode* o_head = odd, * e_head = even;
        
        while(odd && even && even->next){
            odd->next = even->next;
            odd = odd->next;
            
            even->next = odd->next;
            even = even->next;
        }
        // even->next = NULL;可有可无
        odd->next = e_head;
        return o_head;
    }
};