题解 | #链表的奇偶重排#

/**
 * 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;

    }