题解 | #链表的奇偶重排#在校生记录
链表的奇偶重排
https://www.nowcoder.com/practice/02bf49ea45cd486daa031614f9bd6fc3
/** * 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) return nullptr; //奇数位链表,偶数位链表 ListNode *odd = head, *even = head->next; ListNode *next, *preHead;//工作指针 int size = 1;//size变量为head指针最后所指向的链表的位置 while(head->next){ size++; //断开 next = head->next; head->next = next->next; preHead = head; head = next; } //拼接 if(size%2==0){ preHead->next = even; } else{ head->next = even; } return odd; } };