3道经典面试题

[wx1234567]

A. Given a link list of integers, write a procedure PrintReverse() to print out the integers in reverse order. e.g. if mylist is 3, 6, 2, then PrintReverse(mylist) would print out 2 6 3.
Notes:
1. please handle all special cases
2. mylist has to be the same as before after PrintReverse() has run.


typedef struct node
{
int x;
struct node* next;
} Node;

void PrintReverse(Node* mylist)
{
…. Your code goes here …
}

B. What is the time complexity of your procedure?

C. What is the space complexity of your procedure?

[xiaolizi]

void PrintReverse(Node* mylist)
{
int size = 1;
int* buf = (int*)malloc(size);
int id = 0;

while(mylist->next)
{
if(id >= size)
{
size *= 2;
buf = (int*)realloc(buf, size);
}
buf[id++] = mylist->x;
mylist = mylist->next;
}

while(id > 0)
{
print(buf[id--]);
}

free(buf);
}


时间和空间的取舍，写程序总会碰到这样的问题，
抽象出来，生活中又何尝不是这样，时间万物又何尝不是这样

[xiaolizi]

真是不好意思，两个地方要修改一下

void PrintReverse(Node* mylist)
{
int size = 1;
int* buf = 0;
int id = 0;

while(mylist)
{
if(id >= size)
{
size *= 2;
buf = (int*)realloc(buf, size);
}
buf[id++] = mylist->x;
mylist = mylist->next;
}

if(size > 0)
{
while(id >= 0)
{
print(buf[id--]);
}

free(buf);
}
}

[WuYL7812]

push all the Elements into a stack, and then pop out all element, pop one at a time and print them.
time complexity: O(n)
space complexity: O(n)

[WuYL7812]

pop one at a time and print it.

[jussoo]

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#######################xxxxxxxxxx___ | ___xxxxxxxxxx#######################
#############################################################################

[Flood1984]

上面的都是先进栈，然后出栈，
相当与每个元素访问2次，须占用空间n
还有更好一点的吗？
楼下继续