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palantir 面经# JobHunting - 待字闺中

e*n2012-11-05 08:11

1 楼

struct Thing {
int length;
struct Thing *things[];
};
/* Return a deep copy of thing. The object graph formed by the copy
* should have the same structure as the object graph of the original,
* but they should share no references.
*/

l*52012-11-05 08:11

2 楼

应该还是要考虑各种case。。。palantir很喜欢细节。。很想testing的题。。。

a*x2012-11-05 08:11

3 楼

http://www.leetcode.com/2012/05/clone-graph-part-i.html
very similar, I think

【在 e***n 的大作中提到】

: struct Thing {
: int length;
: struct Thing *things[];
: };
: /* Return a deep copy of thing. The object graph formed by the copy
: * should have the same structure as the object graph of the original,
: * but they should share no references.
: */

K*n2012-11-05 08:11

4 楼

牛，这公司理你了啊，他们据说挺拽啊

【在 e***n 的大作中提到】

e*n2012-11-05 08:11

5 楼

不知道阿。。。反正面我的老美开着免提也

【在 K*********n 的大作中提到】

: 牛，这公司理你了啊，他们据说挺拽啊

q*m2012-11-05 08:11

6 楼

先简化一下比如如何 deep copy
struct Thing{
int length,
struct Thing *next;
}
吧，这个 deep copy就相当于重新 create一个 linked list

【在 e***n 的大作中提到】

p*22012-11-05 08:11

7 楼

DFS就可以了吧。

l*i2012-11-05 08:11

8 楼

随便写了一个，大牛指点
Thing* deepCopy(Thing *ptr)
{
if (ptr == NULL) return NULL;
Thing *ans = new Thing;
ans->length = ptr->length;
for (int i=0; ians->things[i] = deepCopy(ptr->things[i]);
return ans;
}

d*g2012-11-05 08:11

9 楼

貌似遇到circle就无限循环了

【在 l***i 的大作中提到】

: 随便写了一个，大牛指点
: Thing* deepCopy(Thing *ptr)
: {
: if (ptr == NULL) return NULL;
: Thing *ans = new Thing;
: ans->length = ptr->length;
: for (int i=0; i: ans->things[i] = deepCopy(ptr->things[i]);
: return ans;
: }

l*i2012-11-05 08:11

10 楼

有道理。再想想。搞个map记录已经见过的node
Thing* deepCopy(Thing *ptr, map &mp)
{
if (ptr == NULL) return NULL;
if (mp.find(ptr) != mp.end()) return mp[ptr];
Thing *ans = new Thing; mp[ptr] = ans;
ans->length = ptr->length;
for (int i=0; ians->things[i] = deepCopy(ptr->things[i]);
return ans;
}

l*i2012-11-05 08:11

11 楼

嗯，刚刚看了leetcode的答案贴，思路差不多