把字符串中*全部移到字符串的头部---要求时间复杂度和空间复杂度

#include <stdio.h>
//把字符串中*全部移到字符串的头部
void moveCharToHead(char *str)
{
if(str == NULL)
return ;
char *pString = str;
char *tmp = NULL;
while(*pString != '\0')
{
if(*pString == '*')
{
tmp = pString;
while(tmp > str)
{
*tmp = *(tmp - 1);
tmp --;
}
*tmp = '*';
str ++;
}
pString ++;
}
}

int main()
{
char str[] = "*th*an*k* yo*u *v*er*y* m**uc*h !";
int len =  sizeof(str) / sizeof(char);
moveCharToHead(str);
printf("%s\n",str);
}

数组中，数值移动问题.

如下：

int A[nSize]，其中隐藏着若干0，其余非0整数，写一个函数int Func(int* A, int nSize)，使A把0移至后面，非0整数移至 

数组前面并保持有序，返回值为原数据中第一个元素为0的下标。(尽可能不使用辅助空间且考虑效率及异常问题，注释规范且给出设计思路)

#include <stdio.h>

int Func(int* A, int nSize)
{
int tag = 0,nCount = 0;
int *p = NULL,*tmp = NULL;
if((A == NULL) || (nSize <= 0))
{
return -1;
}
while(nCount <  nSize)
{
if(A[ nCount ] == 0)
{
tag = 1;
break;
}
else
{
nCount ++;
}
}
if(tag == 1)
{
p = A + nCount;
while(p < A + nSize)
{
if(*p == 0)
{
tmp = p;
tag = *p;
while(tmp < A + nSize)
{
*tmp = *(tmp + 1);
tmp ++;
}
nSize --;
*(A + nSize) = tag;
}
else
{
p ++;
}

}
return nCount;
}
else
{
printf("Donnot find the zero.\n");
return -1;
}
}

int main()
{
int A[] = {0,8,0,7,6,5,0,3,2,0,1,0,8,0,9};

int i,len = sizeof(A) / sizeof(int);

Func(A,len);

for(i = 0;i < len;i ++)
{
printf("%3d",A[i]);
}
printf("\n");
}

好的算法：

void DeleteChars(char* pStrSource, const char* pStrDelete)
{
if(NULL == pStrSource || NULL == pStrDelete)
return;
const unsigned int nTableSize = 256;
int hashTable[nTableSize];
memset(hashTable, 0, sizeof(hashTable));
const char* pTemp = pStrDelete;
while ('\0' != *pTemp)
{
hashTable[*pTemp] = 1;
++ pTemp;
}
char* pSlow = pStrSource;
char* pFast = pStrSource;
while ('\0' != *pFast)
{
if(1 != hashTable[*pFast])
{
*pSlow = *pFast;
++ pSlow;
}
++pFast;
}
*pSlow = '\0';
}