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) {
        // write code here
        if(head == nullptr || head->next == nullptr)
            return head;
        ListNode* odd = head;
        ListNode* even = head->next;


        ListNode* oddHead = head;
        ListNode* evenHead = head->next;

        while (odd->next != nullptr && even->next != nullptr) {
            if (even->next == nullptr) {
                odd->next = nullptr;
            }else {
                odd->next = even->next;
                odd = odd->next;
            }

            if (odd->next == nullptr){
                even->next = nullptr;
            }else {
                even->next = odd->next;
                even = even->next;
            }
        }
        odd->next = evenHead;
        return head;
    }
};