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;  //用两个指针表示要插入的位置

        int index = 3;
        while(even->next){
            if(index%2){
                ListNode *temp = even->next;
                even->next = even->next->next;
                temp->next = odd->next;   //不能是指向even，多轮过后even已经移动
                odd->next = temp;

                odd = odd->next;
            }else{
                even = even->next;
            }
            ++index;
        }

        return head;
    }
};