在线编程题

"ABCDEFGH",8,4

返回："FGHABCDE"

void

rotate(

char

*s1,

int

n,

int

p)

{

for

(

int

i = 0; i < n; i++)

{

s1[i + n] = s1[i];

}

s1[n + n] =

'\0'

;

for

(

int

i = p+1; i < p + n+1; i++)

printf

(

"%c"

, s1[i]);

printf

(

"\n"

);

}

class

StringRotation {

public

:

string rotateString(string A,

int

n,

int

p) {

string B = A + A;

return

B.substr(p+1, n);

}

};

[[1,2,3],

[4,5,6],

[7,8,9]],3

返回：[[7,4,1],

  [8,5,2],

  [9,6,3]]

12345678910111213141516171819
[code]class 
Rotate {
public
:
vector<vector<
int
> > rotateMatrix(vector<vector<
int
> > mat,
int
 
n){
// write code here
vector<
int
> v1;
vector<vector<
int
>> v_ret;
for
 
(
int
 
i = 0; i < n; i++)
{
v1.clear();
for
 
(
int
 
j = 0; j < n; j++)
{
v1.push_back(mat[n - 1 - j][i]);
}
v_ret.push_back(v1);
}
return
 
v_ret;
}
};
[/code]
3.请把纸条竖着放在桌⼦上，然后从纸条的下边向上⽅对折，压出折痕后再展 开。此时有1条折痕，突起的⽅向指向纸条的背⾯，这条折痕叫做“下”折痕 ；突起的⽅向指向纸条正⾯的折痕叫做“上”折痕。如果每次都从下边向上⽅ 对折，对折N次。请从上到下计算出所有折痕的⽅向。给定折的次数n,请返回从上到下的折痕的数组，若为下折痕则对应元素为"down",若为上折痕则为"up".测试样例：3
返回：["down","down","up","down","down","up","up"]


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2

3

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5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21
class
 
FoldPaper

{

public
:

vector<string> foldPaper(
int
 
n)

{

// write code here

vector<string> v;

pushs(v, n,
"down"
);

return
 
v;

}

void
 
pushs(vector<string> &v,
int
 
n,string s)

{

if
 
(n > 0)

{

pushs(v, n - 1,
"down"
);

v.push_back(s);

pushs(v, n - 1,
"up"
);

}

}

};


提示：折痕其实是二叉树结构。该二叉树的特点是：根节点是下，每一个节点的左节点是下，右节点是上。该二叉树的中序遍历即为答案，但不需要构造一颗二叉树，用递归方法可打印出来。