Y********f
发帖数: 410 | 1 测试过的代码：
#include
#include
using namespace std;
void deleteNode(int heap[], int n, int pos)
{
// put the last elements to where the node removed
heap[pos] = heap[n-1];
// adjust elements above pos
int i = pos;
while(i != 0 && heap[i] < heap[(i - 1) / 2])
{
swap(heap[i], heap[(i - 1) / 2]);
i = (i - 1) / 2;
}
// adjust elements below pos
i = pos;
while(true)
{
if (2 * i + 1 > n -2)
{
//al... 阅读全帖 |
|
i**********e
发帖数: 1145 | 2 Already found a bug in your code.
--> int length=sizeof(input)/sizeof(int);
This will always set lenght as 1. Because input is an int* pointer that's
passed into the function, therefore sizeof(input) = sizeof(int*) = 4. |
|
i**********e
发帖数: 1145 | 3 Already found a bug in your code.
--> int length=sizeof(input)/sizeof(int);
This will always set lenght as 1. Because input is an int* pointer that's
passed into the function, therefore sizeof(input) = sizeof(int*) = 4. |
|
q******0
发帖数: 15 | 4 See the code below. I am confused that "len = 5" from main function, "len =
4" from RemoveDuplicate function. When does terminate '\0' count? Thanks.
#include
using namespace std;
void RemoveDuplicate(char str[])
{
int len = sizeof(str)/sizeof(str[0]);
cout << "len = " << len << endl;
}
int main()
{
char a[] = "abcd";
int len = sizeof(a)/sizeof(a[0]);
cout << "len = " << len << endl;
RemoveDuplicate(a);
return 0;
} |
|
i**********e
发帖数: 1145 | 5 Try this and figure out why:
void foo(int str[])
{
int len = sizeof(str)/sizeof(str[0]);
cout << "len = " << len << endl;
}
int main()
{
int a[] = {1,2,3,4};
int len = sizeof(a)/sizeof(a[0]);
cout << "len = " << len << endl;
foo(a);
return 0;
} |
|
w****x
发帖数: 2483 | 6 struct NODE
{
string str;
NODE* pLft;
NODE* pRgt;
NODE(const char* szStr = "") : str(szStr), pLft(NULL), pRgt(NULL) {}
};
void serialize(NODE* pNode, char*& p)
{
if (p == NULL) return;
if (pNode == NULL)
{
*((int*)p) = 0;
p += sizeof(int);
return;
}
int nLen = pNode->str.length();
*((int*)p) = 1;
p += sizeof(int);
strcpy(p, pNode->str.c_str());
p += nLen+1;
serialize(pNode->pLft, p);
serialize(pNode->pRgt... 阅读全帖 |
|
h****n
发帖数: 1093 | 7 我特意跑了一下程序
void foo(int *p1, int val1, int val2)
{
int j;
int *p2, *p3;
int * buf1 = (int*)malloc(sizeof(int)*20);
for(int j =0;j<20;j++)
buf1[j] = 1;
p2 = (int *)buf1;
//val1放到0x12fec8地址也就是当前p2指针指向的位置
*p2 = val1;
//把p1指向val1地址所指向的位置
p1 = (int *)*p2;
for (int i=1; i<10; i++)
{
p2 = (int *)buf1;
*p2 = val1;
p3 = (int *)*p2;
val1+=sizeof(int);
*p3 = val2;
}
printf("buf1:");
for(j =0;j<20;j++)
... 阅读全帖 |
|
w****x
发帖数: 2483 | 8 后来想了一下是不是可以用greedy算法来做，仔细分析了一下发现不可行，果然写了一
个不能得到最优解。
int _inner_greedy(int a[], int n)
{
if (n <= 2)
return 0;
int nMinIndex = 1;
for (int i = 1; i < n-1; i++)
{
if (abs((a[i] - a[0]) - (a[n-1] - a[i])) < abs((a[nMinIndex] - a[0])
- (a[n-1] - a[nMinIndex])))
nMinIndex = i;
}
return a[n-1] - a[0] + _inner_greedy(a, nMinIndex+1) + _inner_greedy(a+
nMinIndex, n-nMinIndex);
}
int getMinCostGreedy(int a[], int n)
{
if (NULL == a || n <= 1)
... 阅读全帖 |
|
w****x
发帖数: 2483 | 9 后来想了一下是不是可以用greedy算法来做，仔细分析了一下发现不可行，果然写了一
个不能得到最优解。
int _inner_greedy(int a[], int n)
{
if (n <= 2)
return 0;
int nMinIndex = 1;
for (int i = 1; i < n-1; i++)
{
if (abs((a[i] - a[0]) - (a[n-1] - a[i])) < abs((a[nMinIndex] - a[0])
- (a[n-1] - a[nMinIndex])))
nMinIndex = i;
}
return a[n-1] - a[0] + _inner_greedy(a, nMinIndex+1) + _inner_greedy(a+
nMinIndex, n-nMinIndex);
}
int getMinCostGreedy(int a[], int n)
{
if (NULL == a || n <= 1)
... 阅读全帖 |
|
j*****y
发帖数: 1071 | 10 如果是 定义 成 int A[10][10]的形式的话
可以用 sizeof(A[0]) / sizeof(int) 得到列数
sizeof(A) /sizeof(int) 得到总数 |
|
b******7
发帖数: 92 | 11 数组作参数要带上长度
void func(int a[],size_t len)
c/c++不能直接用参数传递数组，而将其转化为指针传递
上面的定义等价于void func(int * a,size_t len)
测试如下的case就可知道
void func(int a[], size_t len)
{
cout<
}
int main()
{
int a[] = {1,2};
cout<
func(a,sizeof(a)/sizeof(int));
return 0;
} |
|
r***e
发帖数: 29 | 12 最后个人标准答案
#ifndef _ROMON_HEADER_
#define _ROMON_HEADER_
#define _DEBUG_
#include
#include
//ROMON digits
const std::string ROMON_DIGITS[] = {"I","IV","V","IX", "X","XL","L","XC","C"
,"CD","D","CM","M" };
//ROMON scale
const int ROMON_SCALE[] = {1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900,
1000};
const int ROMON_MAX = 3999;
const int ROMON_MIN = 1;
class RomanNumeralGenerator
{
public:
virtual std::string generator(int num) = 0;
};
class CRoman : public RomanNumeralGene... 阅读全帖 |
|
b******7
发帖数: 92 | 13 #define BLOCK_SIZE 1024
#define ALIGN(size,unit) (((size)+(unit-1))/(unit)*(unit))
struct ChunkHead{
ChunkHead * next;
};
#define CHUNK_SIZE ALIGN(sizeof(ChunkHead) + BLOCK_SIZE, sizeof(int))
ChunkHead * firstChunk;
void initChunk(void * addr, size_t size)
{
firstChunk = NULL;
Chunk * pre = NULL;
for(int i = 0; i < size/CHUNK_SIZE; i++)
{
ChunkHead * cur = (ChunkHead *)(addr + i* CHUNK_SIZE);
if( pre != NULL)
... 阅读全帖 |
|
R*****i
发帖数: 2126 | 14 这个题目建议用红黑树做可能最快捷有效。红黑树每一次插入，删除都是log(k),求平
均值非常简单，就是根部的那个节点。所以总的计算量是O(nlog(k))。
以下是c/c++代码（需要修改数组，以便一个一个输入）。
#include
#include
#include
#define INDENT_STEP 4
enum rbtree_node_color {RED, BLACK};
typedef struct rbtree_node_t {
int value;
struct rbtree_node_t* left;
struct rbtree_node_t* right;
struct rbtree_node_t* parent;
enum rbtree_node_color color;
} *rbtree_node;
typedef rbtree_node node;
typedef enum rbtree_node_color color;
typedef struct... 阅读全帖 |
|
p****o
发帖数: 46 | 15 cool. 什么编程语言？第二题, 上个c++:
using namespace std;
list intersect(list intList1, list intList2){
list::const_iterator it1 = intList1.begin();
list::const_iterator it2 = intList2.begin();
list intList3;
while((it1 != intList1.end()) && (it2 != intList2.end())) {
if (*it1 == *it2) {
intList3.push_back(*it1);
++it1;
++it2;
} else if (*it1 < *it2) {
++it1;
... 阅读全帖 |
|
s******d
发帖数: 424 | 16 5个里面选三个
#include
#include
#include
#include
#include
int main()
{
int values[] = { 1, 2, 3, 4, 5};
int elements[] = { 1, 1, 1, 0, 0};
const size_t N = sizeof(elements)/sizeof(elements[0]);
assert(N == sizeof(values)/sizeof(values[0]));
std::vector selectors(elements, elements + N);
int count = 0;
do
{
std::cout << ++count << ": ";
for (size_t i = 0; i < selectors.size(); ++i)
{
if (selectors[i])
{
... 阅读全帖 |
|
H**********5
发帖数: 2012 | 17 后天一个on site,
刷了3天的C语言，
遇到这个题，感觉很有趣，
写一个统一适用的swap:# define swap(x,y) /
{ void *_tmp=malloc(sizeof(x)); /
void *_x=&x; void *_y=&y; /
memcpy(tmp,_x,sizeof(x));/
memcpy(_x,_y,sizeof(y));/
memcpy(_y,_tmp,sizeof(x));}
这个表面上看上合理，
但一遇到含有指针元素的结构体，感觉就挂了吧，
请问有没有办法实现这种任何任意种类元素的swap，用C实现。 |
|
i*****o
发帖数: 105 | 18 #include
#include
int x[10][10] = {
{ 1, 1, 1, 1, 1, 1, 5, 1, 1, 1,},
{ 1, 1, 1, 1, 1, 1, 4, 1, 1, 1,},
{ 1, 1, 1, 1, 1, 1, 8, 1, 1, 1,},
{ 1, 1, 1, 1, 5, 4, 10, 1, 1, 1,},
{ 1, 1, 1, 1, 4, 1, 10, 10, 1, 1,},
{ 1, 1, 1, 1, 5, 5, 10, 10, 10, 10,},
{ 1, 1, 5, 6, 1, 5, 11, 1, 1, 1,},
{ 1, 1, 3, 1, 1, 1, 10, 1, 1, 1,},
{ 1, 2, 2, 1, 1, 1, 1, 1, 1, 1,},
{ 2, 1, 10, 10,... 阅读全帖 |
|
t*****d
发帖数: 525 | 19 【 以下文字转载自 PDA 讨论区 】
发信人: weidong (伊拉克学习小组副组长), 信区: PDA
标 题: 老码农骂小码农的强文。
发信站: BBS 未名空间站 (Mon Nov 2 19:52:49 2015, 美东)
Linus Tovalds骂瞎写代码的小码农，荡气回肠，应该下发所有码农学习。
Christ people. This is just sh*t
.The conflict I get is due to stupid new gcc header file crap. But what
makes me upset is that the crap is for completely bogus reasons.
This is the old code in net/ipv6/ip6_output.c:
mtu -= hlen + sizeof(struct frag_hdr);
and this is the new “improved” code that uses fancy stuff that wants
magical built-in... 阅读全帖 |
|
a****o
发帖数: 686 | 20 O(N+XlogX) and no extra space is required.
your asymptotic upper limit is either O(N) or O(XlogX), depends on ratio of X/N.
//
#include
#include
#include
#include
#include
using namespace std;
template
void top_X_elements(vector& v, int X){
nth_element( v.begin(),v.begin()+X, v.end(),greater() );//O(N)
sort(v.begin(), v.begin()+X);//O(XlogX)
}
int main(int argc, char* argv[]){
int a[]={1,2,3,4,5,6,7,8,9,10,11,12};
ran... 阅读全帖 |
|
r****y
发帖数: 26819 | 21 not exactly original but can simulate:
/*
* sizeof.c - Implementation of sizeof operator
* http://faq.zanvar.in
*/
#include
#include
/* Find the size of an variable */
#define sizeof_var( var ) ((size_t)(&(var)+1)-(size_t)(&(var)))
/* Find the size of a data type */
#define sizeof_type( type ) (size_t)((type*)1000 + 1 )-(size_t)((type*)1000)
int main ( void )
{
int a;
int b[10];
printf ( "%lu\n", sizeof ( b ) );
printf ( "%lu\n", sizeof ( b+0 ) );
printf ( "%lu\n", sizeof_va |
|
Q**g
发帖数: 183 | 22 whenever you pass an array as an argument to a function call, it is actually
treated as a pointer. The size information is lost to the function.
try this simple one:
#include
using namespace std;
void f(char *array[]){
cout<<"in function, sizeof(array)="<
}
int main(){
char *a[20] = {"abc","def","ghik"};
cout<<"in main, sizeof(a)="<
f(a);
return 0;
} |
|
c*****i
发帖数: 25 | 23 我知道这些问题显得都很笨。请表拍我。谢谢您的指教。
比如说下面这段code
struct a {
char *b;
int c;
};
int n, m, i;
struct a *pa = (struct a*)malloc(n, sizeof(struct a));
for(i = 0; i < n; i++) {
pa[i].b = (char*)malloc(m, sizeof(char));
}
//需要把每一个pa[i].b都free吗？
for(i = 0; i < n; i++) {
free(pa[i].b);
}
//还是只有这句就够了？
free(pa);
还有一段code： ：）
char **s;
int n, m, i;
s = (char**)malloc(n, sizeof(char*));
for(i = 0; i < n; i++) {
s[i] = (char*)malloc(m, sizeof(char));
}
// 需要把每一个s[i]都free吗？
for(i = 0; i < n; i++) {
fre |
|
d*******d
发帖数: 2050 | 24 来自主题: Programming版 - 来，出个题 非常非常简单。
先自己写下结果，然后运行试试。
void f(char a[]){
cout << sizeof(a) << endl;
cout << sizeof(*a) << endl;
}
main(){
char a[10];
cout << sizeof(a) << endl;
cout << sizeof(*a) << endl;
f(a);
} |
|
g*********s
发帖数: 1782 | 25 查了源代码。_Rb_tree_node_base确实有alignment问题。另外Rb_tree还有一个size_type的字段，一个compare字段。
所以sizeof(map<>) = sizeof(_Rb_tree_node_base) + sizeof(size_type) + sizeof(void *) = 32 + 16 = 48.
16。 |
|
s***e
发帖数: 122 | 26 看来是因为这一句：
fread(&sacbin, sizeof(header), 1, fpr);
要改成
fread(&sacbin, sizeof(struct header), 1, fpr);
如果你是用C编译器的话。
当然你也可以定义struct header为typedef struct header {...} HEADER; 那样你就可以用sizeof(HEADER)了。
下面这段代码是可以编译运行的。
#include
struct header {
int hd;
};
struct sac
{
struct header hdr;
float * data;
};
void test2() {
const char * const fn = "test2.txt";
FILE* fpw = fopen(fn, "w");
struct header hdr;
hdr.hd = 18;
fwrite(&hdr, sizeof(hdr), 1, fpw);
fclose(fpw |
|
c**m
发帖数: 30 | 27 that's obvious.
but question asks for difference behind the scene, no mention of anything
specific about memory p1 pointed to and p2 (candidate is supposed to figure
that out). and there're very big differences depending on where they appear
. besides, there's nothing stopping me from making var a pointer to an
integer at location 5:
int var =5;
*((int*)var) = 5;
also, I always thought sizeof(int) == sizeof(void*) but apparently not
because visual studio 64-bit sizeof(int) == 4 and sizeof(voi |
|
i****d
发帖数: 255 | 28 // 定义二维数组的指针
double **zzz；
int i, j, irow = 100， icol = 4；
// 有两种方式malloc
// 第一种
zzz = (double**) malloc( sizeof(double) * irow * icol) );
for(i=0; i
zzz[i] = (double *) malloc( sizeof(double) * icol) );
// 第二种
zzz = (double**) malloc( sizeof(double*) * irow) );
for(i=0; i
zzz[i] = (double *) malloc( sizeof(double) * icol) );
*******************************************
感觉第二种方式好理解：先定义一个指针数组zzz[irow]；然后让里面的各个指针指向
一个个一维数组，即zzz[0]存储第0列的首地址，等等。
但第一种方式不好理解：感觉开辟了两份内存。
请指教！ |
|
l*******G
发帖数: 1191 | 29 can we use the second method to define multiple-dimension arrays (>2)?
i.e. something like
int idim1=100;
int idim2=200;
int idim3=300;
int idim4=10;
zzz = (double**) malloc( sizeof(double*) * idim1) );
for(i=0; i
zzz[i] = (double **) malloc( sizeof(double) * idim2) );
for(j=0; j
zzz[i][j]=(double **) malloc( sizeof(double) * idim3) );
for(k=0; k
zzz[i][j][k]=(double **) malloc( sizeof(double) * idim4) );
}
}
} |
|
v****s
发帖数: 1112 | 30 const int MAX = 500;
char *src = (char*)malloc(MAX*sizeof(char));
for( int i = 0; i
printf("%d\n",sizeof(src));
结果是4，而不是500,而用 char src[MAX] 就可以得到这个正确的sizeof(src)。
char *src 和 char src[] 都是分配在heap上面的么？ sizeof()函数具体是怎么实现
的？
谢谢！！ |
|
y**b
发帖数: 10166 | 31 void func() {
char s[]="abc"; // 在stack上分配空间存储"abc",sizeof(s)等于4
int i[]={1,2,3}; //同上, sizeof(i)等于12
char *ps="def"; // 在常量区存储"def",sizeof(ps)==8
int *pi=new int[3]; // 在heap分配空间存储,sizeof(pi)==8
delete [] pi;
}
以上对否？
这个常量存储区同初始化数据段/非初始化数据段是什么关系？
看网上有人把c/c++内存布局描述如下：
1、栈区（stack）
2、堆区（heap）
3、全局区（静态区）：初始化数据段/非初始化数据段
4、文字常量区
5、程序代码区 |
|
l***y
发帖数: 4671 | 32 结果：
sizeof(int(*)[1000][1000][1000]): 8
sizeof(double(*)[1000][1000][1000]): 8
sizeof (int(*)): 8
sizeof (double(*)): 8 |
|
t*******s
发帖数: 7 | 33 #include
#include // malloc ,free
#include //memcpy
typedef double data_t ;
typedef struct {
size_t dims; //number of dimensions
size_t *dim_define; //dim1=10, dim2=10, dim3=20, etc.
data_t *data; //one dimension storage
size_t size ;//size of (data) -optional
}matrix_t ;
matrix_t * matrix_new( size_t dims, size_t *dim_define)
{
size_t i, size=1;
for(i=0; i
size*=dim_define[i];
}
matrix_t *ret=malloc( siz... 阅读全帖 |
|
d****n
发帖数: 233 | 34 I think the optimization in Kwn's solution could have bugs. Here is mine
without any optimization:
int FindSubSeqWithMaxMinDistance(int *X, int N, int K)
{
if ( N < 2 || K < 2 || K > N ) return 0;
// Alloc buffers
int **F = (int **)malloc(K*sizeof(int *)+ K*N*sizeof(int));
int *temp = (int *)(F + K);
for (int i = 0; i < K; i++)
{
F[i] = temp + N*i;
}
memset(temp, 0, K*N*sizeof(int));
for(int j = 0; j < N; j++)
{
F[0][j] = INT_MAX;
}
// Assume A |
|
w******1
发帖数: 520 | 35 写函数实现一个数组找2个数的和是一个固定值。
code：bool has_num(int *integers, int target)
{
for (int i=0;i
{
if(num_hashtable(target-*(integers+i))
{
return true;
}
}
return false;
}
====================
1. 边界条件是否需要做个判断呢？ 比如*integers 如果是NULL? int target 如果长
度小于等于0？
2.
sizeof(char *) = 4 bytes 或者2 BYTES 或者其他数字，根据不同的编译环境。
sizeof(integers) 可能得到的结果不是你想要的， 结果也许就是指针的SIZE =4 |
|
h*****g
发帖数: 944 | 36 C++里是不是现在没有现成的get array size/length的function??
如果没有的话，我在网上看到了一个code
#include
template char (&array(T(&)[N]))[N];
#define length(a) (sizeof a / sizeof a[0])
int main()
{
int a[10];
std::cout << sizeof array(a) << '\n';
std::cout << length(a) << '\n';
}
请问这个code对所有的data type都work吗？ 我试试了，好象是可以的。中间有什么错
误吗？ |
|
c**********e
发帖数: 2007 | 37 #include
class A {
public: virtual ~A() {}
};
class B : public A {
public:
virtual void f() {}
};
Referring to the sample code above, on a typical platform that uses virtual
tables to implement virtual function lookup, what do you expect "sizeof(B)"
to be?
a) 4
b) sizeof(void *)
c) 2*sizeof(void *) |
|
a****n
发帖数: 1887 | 38 5.
memset (str,0,sizeof(str));
memset (str,1,sizeof(str));
memset (str,0xff00,sizeof(str));
有没有问题？
这三个语句初始化后的内存是什么样子的 |
|
w**********y
发帖数: 1691 | 39 ////////////////////////////
class A{ }; A a;
那么 sizeof(a) is 1;
class X{
void doNothing(){}
char a;
};
sizeof(x) 咋还是1,不是 1+1 =2 呢?
////////////////////////////
class Y{
virtual void doNothing(){}
};
sizeof(y) is 4..这个应该是指向virtual table的地址大小吧?
class Y{
virtual void doNothing(){}
virtual void doNothing2(){}
char a;
};
为啥是8不是4+1=5啊? 而且再添加一个member: char b..还是8..
如果换成char* b就成了12...
如果换成char a[2],结果是8
如果换成char a[10],结果是16
非常不理解.. |
|
z****o
发帖数: 78 | 40 存储city之间的距离的题目，可以有一下两个假设：
1. city1 到 city2 = city2 到 city1
2. city1 到 city2 距离是非负的。
这样的话，可以计算 cityi 到 cityj 的最短路径的边集S(i,j) = { e1,e2,e3..}
S = S(0,1) U S(0,2) U S(0,3) ..... U S(1,2) U S(1,3) U ..... U S(i,i+1) U S
(i,i+2) U ...
最后存下来S就可以了。这种存法的空间是 |S|*( 2*sizeof(int) + 1* sizeof(double
) );
这个空间去和 ((n-1)+(n-2)+(n-3)+...+2+1)*sizeof(double) 的矩阵存法比较一下，用
空间使用比较省的一种来存。|S|够小的时候用第一种比较好，比如所有的city在一条
线上。
|S|不小的时候用第二种比较省，比如city均匀分布在圆上。 |
|
a***r
发帖数: 93 | 41 memory O(n)
time O(nlogn)
//~ Given a array,find out if there exist a subarray such its sum is zero
#include
#include
using namespace std;
static void swap(int *p, int i, int j) {
int t=p[i]; p[i]=p[j];p[j]=t;
}
static int partition(int *p,int left,int right) {
int j=left;
for(int i=left;i
if(p[i]
}
swap(p,j,right);
return j;
}
static void binary_sort(int *p, int left, int right) {
if(left>=r... 阅读全帖 |
|
a******u
发帖数: 239 | 42 I think my answers to Question 7 and 8 are correct, but they said they are
not correct, why?
I already tested the code by Visual C++ (c compiler).
Thank you very much.
Line in file Contents
30 int * someIDs, theFirst, *r;
110 someIDs =GetSomeIDs(); /* defined below */
111 theFirst = someIDs [0];
112 r= ReorderIDs(someIDs);
113-150 /* we want to use ‘theFirst’ and ‘r’ here*/
499 /*-------- GetSomeIDs-----*/
500 int * GetSomeIDs()
501 {
502 int ids[8];
503-5... 阅读全帖 |
|
r*****k
发帖数: 1281 | 43 void RemoveDuplicate(char str[])
{
int len = sizeof(str)/sizeof(str[0]);
cout << "len = " << len << endl;
}
这上面这段code中，sizeof(str)是计算“the size of the pointer represented by
the array identifier”, which is always 4. |
|
a**U
发帖数: 115 | 44 我也打印出来，sizeof(A)=4, sizeof(B)=4, sizeof(C) = 8, 我用的ubuntun, 编译是
g++ cast.cpp 出来 a.out. 运行a.out，没有问题。奇怪的很！ |
|
g****e
发帖数: 172 | 45 char *words[] = {"stately", "plump", "buck", "mulligan"};
// calculate how many elements in words
size_t words_size = sizeof(words)/sizeof(char *);
为什么不是直接 size_t words_size = sizeof(words)? |
|
d*******u
发帖数: 186 | 46 From c programming language:
typedef long Align; /* for alignment to long boundary */
union header { /* block header */
struct {
union header *ptr; /* next block if on free list */
unsigned size; /* size of this block */
} s;
//The Align field is never used; it just forces each header to be
aligned on a worst-case boundary.
Align x; /* force alignment of blocks */
};
typedef union header Header;
static Header base; /* empty list to get started */
static Header *fr... 阅读全帖 |
|
f*****7
发帖数: 92 | 47 我个人不喜欢swap的方法，自己没法理解
排序，然后跳过重复元素
希望能帮上你
class Solution {
public:
vector > permuteUnique(vector &num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
sort(num.begin(), num.end());
int len = num.size();
int* path = (int*) malloc(sizeof(int) * len);
bool* visited = (bool*) malloc(sizeof(int) * len);
memset(visited, false, sizeof(bool) * len);
vector > bigList;
int... 阅读全帖 |
|
b*******7
发帖数: 907 | 48 1. what's the output? why?
#include
#include
int main()
{
while(1)
{
fprintf(stdout,"hello-std-out");
fprintf(stderr,"hello-std-err");
sleep(1);
}
return 0;
}
2. what's the output?
#include
int main()
{
float a = 12.5;
printf("%d\n", a);
printf("%d\n", (int)a);
printf("%d\n", *(int *)&a);
return 0;
}
3. can it be built? if yes, what's the result?
file1.c:
int arr[80];
file2.c:
extern int *arr;
int main(... 阅读全帖 |
|
r*******n
发帖数: 3020 | 49 可以，
result = []
1 left, right = partition // the elements in right > that of left
2 if sizeof(right) > k
drop the left, recursively call on the right.
else if sizeof(right) < k
k = k - sizeof(right)
append the right to the result.
recursively call on the left.
else
append the right to the result.
the end.
|
|
