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;
* };
*/
/**
* 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
*/
struct ListNode* oddEvenList(struct ListNode* head ) {
// write code here
if(!head||!head->next)return head;
struct ListNode* yhead1=(struct ListNode*)malloc(sizeof(struct ListNode));
struct ListNode* yhead2=(struct ListNode*)malloc(sizeof(struct ListNode));
yhead1->next=head;
yhead2->next=head->next;
struct ListNode* odd = yhead1->next;
struct ListNode* even=yhead2->next;
while (even&&even->next) {
odd->next=odd->next->next;
odd=odd->next;
even->next=even->next->next;
even=even->next;
}
odd->next=yhead2->next;
return yhead1->next;
}
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