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* p1 = head;
        ListNode* even = head->next;
        ListNode* p2 = even;
        while (p2&&p2->next) {
            p1->next = p2->next;
            p1 = p1->next;
            p2->next = p1->next;
            p2 = p2->next;
        }
        p1->next = even;
        return head;
    }
};