求方法！将一个数组的前k个数和后n-k个数交换，要求时间O(n),空间O(1)???

[dxrenderman]

求方法！将一个数组的前k个数和后n-k个数交换，要求时间O(n),空间O(1)???

[xiaocai0001]

n > 2k ?

[dxrenderman]

no,比如a[6]={1,2,3,4,5,6};
k=2,交换后就为3,4,5,6,1,2
k=5,交换后就为6,1,2,3,4,5

[shootsky]

for(i=k;i!=0;i--)
{
temp=a[k-1];
a[k-1]=a[n-1];
a[n-1]=temp;
temp=a[k-1];
a[k-1]=a[n-1-k];
a[n-1-k]=temp;


}

处女回帖，没有调试，不知对不对

[cunsh]

好像不对呀.处女.

[shootsky]

发得太仓促，错了，再改改

[shootsky]

不知道用冒泡法行不行，高手教教我

a[0]冒到a[n-k]
a[1]冒到a[n-k+1]
a[k]冒到a[n]

哎，我还太嫩了

[xiaocai0001]

#include <iostream>

void SwapN(int array[], int nLen, int beginPos);

int main()
{
int array[] = {1,2,3,4,5,6,7,8,9};
int len = sizeof(array)/sizeof(int);
int i;
for(i=0; i<len; ++i)
std::cout<<array[i]<<" ";
std::cout<<std::endl;
SwapN(array, sizeof(array)/sizeof(int), 3);
for(i=0; i<len; ++i)
std::cout<<array[i]<<" ";
std::cout<<std::endl;
return 0;
}

// int array[] : 待操作的数组
// int nLen : 数组长度
// int k : 起始位置
void SwapN(int array[], int nLen, int k)
{
k = k % nLen;
if(k == 0)
return;
if(k < 0)
k += nLen;

int count = 0;
int index = 0;
int initpos = -1;
while(count < nLen)
{
++initpos;
int temp = array[initpos];
for(index=initpos; (index+k)%nLen!=initpos; index=(index+k)%nLen)
{
array[index] = array[(index+k)%nLen];
++count;
}
array[nLen + initpos - k] = temp;
++count;
}
}