2019年校招：美光半导体笔试题目

2019年美光半导体笔试题目

1、What gets printed()(5’)

counter = 1
def doLotsOfStuff():
	global counter
	for i in (1,2,3):
		counter+=1
doLotsStuff()
print counter
A.1   B.3   *C.4   D.7   E.none of the above

2、What gets printed()(5’)

numbers = [1,2,3,4]
numbers.append([5,6,7,8])
numbers.extend([9,10])
print len(numbers)
A.4  B.3  *C.7   D.10  E.6  F.An exception is throw

3、What gets printed()(5’)

key2 = {'1':1,'2':[2,4]}
TheCopy = kvps.copy()
kvps['1'] = 5
kvps['2'][1] = 8
sum = kvps['2'][1]+theCopy['2'][1]+theCopy['1']
print sum
A.21  B.13  *C.17  D.15  E.An exception is throw

4、What gets printed()(5’)

class A(object):
	x = 100
	def __init__(self,value):
		self.x = value
a = A(1000)
print a.x+A.x
A.200  *B.1100 C.2000 D.An exception is throw

5、What gets printed()(5’)

def add(a,b):
	return a+b
x = 0
try:
	add('a',1)
except TypeError:
	x |=1<<7
else:
	x |=1<<6
finally hex(x)
*A.0x81 B.0x1 C.0xc1 D.0x80 E.6 F.Anexception is thrown

6、给定一个整型变量a，写出三个表达式，第一个设置a的第四个bit(bit3)为1，第二个清除a的第四个bit(bit3)为0.第三个将a的第四个bit（bit3）取反。在以上操作中，要保持其他位不变(10’)。

a = a&(1<<3)
a = a&(~(1<<3))
a = 

7、Given following C cod, please design proper test flow to cover as many brances as possible(15’).

int GetMaxDay(int year,int month)
{
	int maxday = 0;
	if(month>=1 && month<=12)
	{
		if(month == 2)
		{
			if(year%4 == 0)
			{
				if(year%100 == 0)
				{
					if(year%400 == 0)
						maxday = 29;
					else
						maxday = 28;
				else
					maxday = 29;
			else
				maxday = 29;
		else
		{
			if(month == 4||month == 6||month==9||month == 11)
				maxday = 30;
			else
				maxday = 31;
		}
	}
	return maxday;
}
//参考程序员面试宝典 第五版  269页解法

8、用宏定义实现swap(x,y)，即交换两个数(15’)。

#define swap(a,b) a = a+b,b=a-b,a=a-b

9、static局部变量和普通局部变量的区别(15’)。

Example:
int example(int a,int b)
{
	static int c =0;
	c = a+b;
	return c;
}
//提示，在调用后是否会被清空、下一次调用值是否会保留上一次的结果

10、编写C/C++程序实现如何找出单链表中的倒数第K个元素（20’）

//提示：两个指针距离为K，后面一个指针指向尾部，另外一个指针所指即为所求
//参考：
/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
            val(x), next(NULL) {
    }
};*/
class Solution {
public:
    ListNode* FindKthToTail(ListNode* pListHead, unsigned int k)
    {
        ListNode *p1,*p2;
        p1 = pListHead;
        p2 = pListHead;
        if(pListHead == NULL||k <= 0)
            return NULL;
        for(int i = 0;inext!=NULL)
                p2 = p2->next;
            else
                return NULL;
        }
        while(p2->next!=NULL)
        {
            p1 = p1->next;
            p2 = p2->next;
        }
        return p1;
    }
};