习题11-7 奇数值结点链表

习题11-7 奇数值结点链表 (20 分)

实现代码：

struct ListNode *readlist(void) {
    struct ListNode *head = NULL;
    struct ListNode *end = NULL;
    int number;
    scanf("%d",&number);
    do {
        struct ListNode *p = (struct ListNode*) malloc (sizeof(struct ListNode));
        if(p != NULL) {
            p->data = number;   //节点创建成功,则保存数据
            p->next = NULL;
        }
        else exit(1);
        if(head != NULL) {      //头节点不为空，说明链表不为空，则在链表尾添加新节点
            end->next = p;
        }
        else {
            head = p;           //第一次运行一次
        }
        end = p;                //新创建的节点必为尾节点
        scanf("%d",&number);
    }while(number != -1);
    return head;
}


struct ListNode *getodd( struct ListNode **L ) {
    struct ListNode *head_1=NULL,*head_2=NULL,*end_1=NULL,*end_2=NULL;
    while((*L) != NULL) {
        int number;
        number = (*L)->data;
        struct ListNode *p = (struct ListNode*)malloc(sizeof(struct ListNode));
        if(number % 2 != 0) {       //为奇数
            if(p != NULL) {
                p->data = number;
                p->next = NULL;
            }
            else exit(1);
            if(head_1 != NULL) {
                end_1->next = p;
            }
            else {
                head_1 = p;
            }
            end_1 = p;
        }
        else {                     //偶数
            if(p != NULL) {
                p->data = number;
                p->next = NULL;
            }
            else exit(1);
            if(head_2 != NULL) {
                end_2->next = p;
            }
            else {
                head_2 = p;
            }
            end_2 = p;
        }
        *L = (*L)->next;          //依次往后判断直到结束
    }
    *L = head_2;
    return head_1;
}