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) {
        // write code here
        if (head == nullptr || head->next == nullptr) {
            return head;
        }
        ListNode* oddList = new ListNode(head->val);
        ListNode* oddtmp = oddList;
        cout << "head->val=" << head->val << endl;

        ListNode* evenList = new ListNode(head->next->val);
        ListNode* eventmp = evenList;
        cout << "head->next->val=" << head->next->val << endl;
        ListNode* tmp = head->next->next;
        int i = 1;
        while (tmp != nullptr) {
            oddtmp->next = new ListNode(tmp->val);
            cout << "oddList->next=" << tmp->val << endl;
            oddtmp = oddtmp->next;
            tmp = tmp->next;
            if (tmp != nullptr) {
                eventmp->next = new ListNode(tmp->val);
                eventmp = eventmp->next;
                cout << "evenList->next=" << tmp->val << endl;
                tmp = tmp->next;
            }
        }

        oddtmp->next = evenList;
        return oddList;
    }
};