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 || head->next->next==nullptr) {
            return head;
        }

        ListNode* fir = head;
        ListNode* index=head->next;
        ListNode* sec = head->next;
        while(fir && fir->next && fir->next->next) {
            fir->next = sec->next;
            fir=fir->next;
            sec->next=fir->next;
            sec=sec->next;
        }
        fir->next = index;
        return head;

    }
};