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 == nullptr || head->next == nullptr) {
            return head;
        }
        ListNode tmp1(0), tmp2(0);
        ListNode *cur = head, *odd = &tmp1, *even = &tmp2;
        bool isOdd = true;
        while (cur) {
            if (isOdd) {
                odd->next = cur;
                odd = odd->next;
            } else {
                even->next = cur;
                even = even->next;
            }
            cur = cur->next;
            isOdd = !isOdd;
        }
        odd->next = tmp2.next;
        even->next = nullptr;
        return tmp1.next;
    }
};