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 || !head->next) return head;
        ListNode* odd = head;
        ListNode* even = head->next;
        ListNode* evenHead = even;
        while (even && even->next) {
            odd->next = even->next;
            odd = odd->next;
            even->next = odd->next;
            even = even->next;
            }
            odd->next = evenHead;
            return head;
        }
};