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* odd = new ListNode(-1), *even = new ListNode(-1);
        ListNode* ot = odd, *et = even;
        int num = 0;
        while(head){
            num++;
            if(num&1){
                ot->next = head;
                ot = head, head = head->next;
                ot->next = NULL;
            }else{
                et->next = head;
                et = head, head = head->next;
                et->next = NULL;
            }
        }
        
        if(odd->next)
            ot->next = even->next;
        return odd->next;
    }
};