美团网笔试题目

美团网笔试题目 解答：是不是题目不完整啊，我算的是 3：1 2、一个汽车公司的产品，甲厂占 40%，乙厂占 60%，甲的次品率是 1%，乙的次品率是 2%，现在抽出一件汽车时次品，问是甲生产的可能性 解答：典型的贝叶斯公式，p(甲|废品) = p(甲 废品) / p(废品) = (0.4 0.01) /(0.4 0.01 + 0.6 0.02) = 0.25 3、k 链表翻转，美团网笔试题目。给出一个链表和一个数k，比如链表 123456，k=2，则翻转后 214365，若 k=3,翻转后321654，若 k=4，翻转后 432156，用程序实现 非递归可运行代码： #include #include #include typedef struct node { struct node *next; int data; } node; void createList(node **head, int data) { node *pre, *cur, *new; pre = NULL; cur = *head; while (cur != NULL) { pre = cur; cur = curnext; } new = (node *)malloc(sizeof(node)); newdata = data; newnext = cur; if (pre == NULL) *head = new; else prenext = new; } void printLink(node *head) { while (headnext != NULL) { printf(%d , headdata); head = headnext; } printf(%d\n, headdata); } int linkLen(node *head) { int len = 0; while (head != NULL) { len ++; head = headnext; } return len; } node* reverseK(node *head, int k) { int i, len, time, now; len = linkLen(head); if (len k) { return head; } else { time = len / k; } node *newhead, *prev, *next, *old, *tail; for (now = 0, tail = NULL; now time; now ++) { old = head; for (i = 0, prev = NULL; i k; i ++) { next = headnext; headnext = prev; prev = head; head = next; } if (now == 0) { newhead = prev; } oldnext = head; if (tail != NULL) { tailnext = prev; } tail = old; } if (head != NULL) { tailnext = head; } return newhead; } int main(void) { int i, n, k, data; node *head, *newhead; while (scanf(%d %d, n, k) != EOF) { for (i = 0, head = NULL; i n; i ++) { scanf(%d, data); createList(head, data); } printLink(head); newhead = reverseK(head, k); printLink(newhead); } return 0; }