题解 | 链表的奇偶重排

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    ListNode* oddEvenList(ListNode* head) {
        if(!head) {
            return nullptr;
        }
        // assert(head);
        ListNode *odd_head = head, *even_head = head->next;
        ListNode *p = head, *q, *last_p = nullptr;
        int len = 1;
        while(q = p->next) {
            p->next = q->next;
            last_p = p;
            p = q;
            ++ len;
        }
        if(len % 2) {
            p->next = even_head;
        } else {
            last_p->next = even_head;
        }

        

        return odd_head;
    }
};