常见的字符串操作

　　字符串（String），是由零个或多个字符组成的有限串行。一般记为 ()。它是编程语言中表示文本的数据类型。

　　通常以串的整体作为操作对象，如：在串中查找某个子串、求取一个子串、在串的某个位置上插入一个子串以及删除一个子串等。

　　两个字符串相等的充要条件是：长度相等，并且各个对应位置上的字符都相等。

　　设p、q是两个串，求q在p中首次出现的位置的运算叫做模式匹配。

　　串的两种最基本的存储方式是顺序存储方式和链接存储方式。

 

//************************************
// Method:    ItoA
// Description:  Transfer an int to a string
// Access:    public 
// Returns:   void
// Qualifier:
// Parameter: int n
// Parameter: char * str
//************************************
void ItoA(int n, char *str)
{
    char num[256];
    int k = n, i;
    if (n<0) k = -n;
        
    for(i=0; k>0; i++)
    {
        num[i] = k%10 + '0';
        k /= 10;
    }


    if(n<0) num[i] = '-';
    else --i;

    int j;
    for(j=0;i>=0;j++,i--)
    {
        str[j] = num[i];
    }
    str[j] ='\0';
}

 

//************************************
// Method:    AtoI
// Description:  Transfer a string to a long integer
// Access:    public 
// Returns:   void
// Qualifier:
// Parameter: long * n
// Parameter: char * str
//************************************
void AtoI(long *n, char* str)
{
    if(str == NULL) throw "error";

    bool sign = false;
    if(*str == '-')
    {
        str++;
        sign = true;
    }
    else if (*str == '+')
    {
        str++;
    }

    *n = 0;
    while(*str)
    {
        if(*str >= '0' && *str <='9')
        {
            *n =(*n)*10 + ((*str) - '0') ;
            str++;
        }
        else
        {
            throw "error";
        }
    }
    if(sign == true){(*n) = -(*n);}
}

 

//************************************
// Method:    LoopMove
// Description:  Move the chars from the string recurrently
// Access:    public 
// Returns:   void
// Qualifier:
// Parameter: char * pStr
// Parameter: int steps: The steps to be moved
//************************************
void LoopMove(char* pStr, int steps)
{
    if (pStr == NULL)
    {
        throw "error";
    }

    int len = strlen(pStr);

    if (steps%len != 0)
    {
        char* snippet = new char[len];
        strcpy(snippet, pStr + len - steps%len);
        *(pStr + len - steps%len) = '\0';
        strcpy(snippet + steps%len, pStr);
        strcpy(pStr, snippet);
    }
}

 

//************************************
// Method:    strCpy
// Description:  Copy string from source to destination
// Access:    public 
// Returns:   char*: Start address of the destination string
// Qualifier:
// Parameter: const char * Src: Source string
// Parameter: char * Dst: Destination string
//************************************
char* strCpy(const char* Src, char* Dst)
{
    if(Src == NULL || Dst == NULL) throw "error";
    char* DstCpy = Dst;
    while((*Dst++ = *Src++) != '\0');
    return DstCpy;
}

 

//************************************
// Method:    FindSubStr
// FullName:  Locate substring. Returns a pointer to the first occurrence of str2 in str1, or a null pointer if str2 is not part of str1. The matching process does not include the terminating null-characters.
// Access:    public 
// Returns:   const char*: A pointer to the first occurrence in str1 of any of the entire sequence of characters specified in str2, or a null pointer if the sequence is not present in str1.
// Qualifier:
// Parameter: const char * str1: C string to be scanned.
// Parameter: const char * str2: C string containing the sequence of characters to match.
//************************************
const char* FindSubStr(const char* str1, const char*str2)
{
    if (str1 == NULL || str2 == NULL)
    {
        throw "error";
    }
    int len1 = 0;
    int len2 = 0;
    const char*p = str1;
    while(*p++) len1++;
    p = str2;
    while(*p++) len2++;
    if (len1 < len2)
    {
        return NULL;
    }

    int i = 0;
    for (;i <= len1 - len2;i++)
    {
        int j = 0;
        while(*(str2 + j))
        {
            if(*(str1 + i + j) == *(str2 + j))
            {
                j++;
            }
            else
            {
                break;
            }
        }
        if (j == len2)
        {
            return (str1 + i);
        }
    }
    return NULL;
}

 

//************************************
// Method:    InverseSentence
// FullName:  Inverse a Sentence which is ended with a '.', while keep the order of each word
// Access:    public 
// Returns:   void
// Qualifier:
// Parameter: char * str
// Parameter: char * res
//************************************
void InverseSentence(char* str, char* res)
{
    int    len = strlen(str);
    int i = len - 1;

    int cnt = 0;//number of letter which is not space
    int cnt_s = 0;//number of letter which is space
    for (;i >= 0;i--)
    {

        if (*(str + i) != ' ' && *(str + i) != '.')
        {
            cnt++;
            if (i == 0 || *(str + i - 1) == ' ' || *(str + i - 1) == '.')
            {
                memcpy(res + len - i - cnt, str + i, cnt);        
                cnt = 0;
            }
        }
        else
        {
            cnt_s++;
            if (*(str + i - 1) != ' ' && *(str + i - 1) != '.')
            {
                memcpy(res + len - i - cnt_s, str + i, cnt_s);        
                cnt_s = 0;
            }
        }
    }
    *(res + len) = '\0';

    //move the dot to the end
    memcpy(res, res + 1, len - 1);
    *(res + len - 1) = '.';
    *(res + len) = '\0';

}