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:
    ListNode* oddEvenList(ListNode* head) {
        if (head == nullptr) return nullptr;
        ListNode* oddList1 = new ListNode(0);
        ListNode* oddList = oddList1;
        ListNode* evenList1 = new ListNode(0);
        ListNode* evenList = evenList1;
        int count = 0;
        while (head) {
            if (count%2 == 0) {
                oddList->next = new ListNode(head->val);
                oddList = oddList->next;
            } else {
                evenList->next = new ListNode(head->val);
                evenList = evenList->next;
            }
            count++;
            head = head->next;
        }
        oddList->next = evenList1->next;
        return oddList1->next;
        
    }
};