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
        ListNode* qi = new ListNode(0);
        ListNode* ou = new ListNode(0);
        
        ListNode* a = qi;
        ListNode* b = ou;
        ListNode* cur = head;
        int index = 1;
        while(cur != nullptr){
            if(index %2 != 0){
                a->next = cur;
                a = cur;
                index++;
            }
            else{
                b->next = cur;
                b = cur;
                index++;
            }
            cur = cur->next;
        }
        b->next = nullptr;
        a->next = ou->next;
        return qi->next;
    }
};