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* JHead = new ListNode(head->val);
        ListNode* OHead = new ListNode(head->next->val);
        
        ListNode* temp = JHead;
        ListNode* tempHead = head;


        while(tempHead->next && tempHead->next->next){

            temp->next = new ListNode(tempHead->next->next->val);
            tempHead = tempHead->next->next;
            temp = temp->next;
        }


        tempHead = head->next;
        temp = OHead;

        while(tempHead->next && tempHead->next->next){

            temp->next = new ListNode(tempHead->next->next->val);
            tempHead = tempHead->next->next;
             temp = temp->next;
        }

        temp = JHead;
        while (temp->next) {
            temp = temp->next;
        }

        temp->next = OHead;
        return JHead;


    }
};