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* oddHead = new ListNode(0);
        ListNode* evenHead = new ListNode(0);
        ListNode* odd = oddHead, *even = evenHead, *curr = head; # 创建头结点是为了后面的拼接
        int count = 1;  // 通过计数判断是奇数位还是偶数位

        while (curr) {  
            if (count % 2 == 1) {  // 奇数位
                odd->next = curr;
                odd = odd->next;
            } else {  // 偶数位
                even->next = curr;
                even = even->next;
            }  
            curr = curr->next;
            count++;
        }

        even->next = nullptr; // 确保偶数链表尾部为null
        odd->next = evenHead->next; // 连接奇数链表和偶数链表

        ListNode* result = oddHead->next;
        delete oddHead; // 清理临时节点
        delete evenHead;
        return result;
    }
};