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 == NULL || head->next ==NULL) return head;
    struct ListNode* oh = head, *eh = head->next, *o = oh, *e = eh;
    while(o->next != NULL && e->next != NULL) {
        o->next = e->next;
        o = o->next;
        e->next = o->next;
        e = e->next;        
    }
    o->next = eh;
    return oh;
}