328.奇偶链表

😍双指针

class Solution {
public:
    ListNode* oddEvenList(ListNode* head) {
        if(!head || !head->next)
            return head;
        ListNode* head2 = head->next, *h2 = head2;
        ListNode* iter = head2->next, *h1 = head;
        bool isEven = true;
        while(iter) {
            if(isEven) {
                h1->next = iter;
                h1 = h1->next;
            } else {
                h2->next = iter;
                h2 = h2->next;
            }
            isEven = !isEven;
            iter = iter->next;
        }
        h1->next = head2;
        h2->next = nullptr;
        return head;
    }
};

Accepted

71/71 cases passed (8 ms)

Your runtime beats 98.43 % of cpp submissions

Your memory usage beats 17.56 % of cpp submissions (9.7 MB)