题解 | #链表的奇偶重排#在校生记录

/**
 * 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) return nullptr;
        //奇数位链表，偶数位链表
        ListNode *odd = head, *even = head->next;
        ListNode *next, *preHead;//工作指针
        int size = 1;//size变量为head指针最后所指向的链表的位置
        while(head->next){
            size++;
            //断开
            next = head->next;

            head->next = next->next;
            preHead = head;
            head = next;
        }

        //拼接
        if(size%2==0){
            preHead->next = even;
        }
        else{
            head->next = even;
        }

        return odd;
    }
};