题解 | #链表的奇偶重排#
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) {
if (!head) {
return nullptr;
}
ListNode *odd = head;
ListNode *even = head->next;
ListNode *pre;
ListNode *evenhead = even;
ListNode *oddhead = odd;
while (odd&&even) {
pre = odd;
odd->next = even->next;
odd = even->next;
if (odd) {
even->next = odd->next;
even = odd->next;
}
}
if (odd) {
odd->next = evenhead;
}
else{
pre->next = evenhead;
}
return head;
// write code here
}
};
if (odd) {
even->next = odd->next;
even = odd->next;
}主要odd为空值的情况
