题解 | #链表的奇偶重排# 饿了效率就是很低。。外卖铃声吵醒对面小哥，很抱歉

/**
 * 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 || !head->next) {
            return head;
        }
        ListNode *oddList = head, *evenList = head->next;
        ListNode *oddHead = oddList, *evenHead = evenList;
        
        while(oddList->next && evenList->next) {
            oddList->next = oddList->next->next;
            evenList->next = evenList->next->next;
            
            oddList = oddList->next;
            evenList = evenList->next;
        }

        oddList->next = evenHead;
        
        return oddHead;
        // write code here
    }
};