数字与字符串相关问题

//1-n整数中1出现的次数:时间复杂度nlog(n),因为数字n的位数为O(log n)
int count1OfNumber(int n){
    int count = 0;
    while (n){
        if (n % 10 == 1)
            count++;
        n /= 10;
    }
    return count;
}
int NumbersOf1Between1AndN(int n){
    if (n < 1){
        g_InvalidInput = true;
        return 0;
    }
    int count = 0;
    for (int i = 1; i <= n; ++i){
        count += count1OfNumber(i);
    }
    return count;
}

//利用hash表实现快速查找：数组实现的hash表O(1)时间查找，STL map实现的hash表O(log n)时间的查找， map适合数据量不大的情况，内部用二叉查找树实现，按照键排序的；
//定义一个函数：输入两个字符串，从第一个字符串中删除在第二个字符串中出现过的所有字符
void deleteCharInString2(string& string1, string& string2){
    if (string1.length() == 0 || string2.length() == 0)
        return;
    int toBeDeleted[256];
    for (int i = 0; i < 256; i++)
        toBeDeleted[i] = 0;
    for (int i = 0; i < string2.length(); ++i){
        toBeDeleted[string2[i]] = 1;
    }
    for (int i = 0; i < string1.length(); ++i) {
        if (toBeDeleted[string1[i]] > 0)
            string1.erase(i, 1);
    }
}

//把字符串转换成整数：
// 1：考虑特殊输入，空指针，空字符串，只有正负号之后没有数字，字符串中有非数字字符
// 2：考虑整型的（正数）上溢出，（负数）下溢出, 需要把num类型设置为long long才能不溢出，再用num与int的极大极小值比较
// 3：输入测试：正数，负数，0
// 注意！使用全局变量来标记非法输入
enum Status {kValid = 0, kInvalid};
Status g_nStatus = kValid;
int StrToIntCore(char* pStr, bool minus){
    long long num = 0;
    while (*pStr != '\0'){
        if (*pStr >= '0' && *pStr <= '9'){
            int flag = minus ? -1 : 1;
            int digit = *pStr - '0';
            num = num * 10 + digit * flag;
            if ((!minus && num > 0x7fffffff) || (minus && num < (signed int)0x80000000)){
                num = 0;
                break;
            }
            pStr++;
        }
        else {
            num = 0;
            break;
        }
    }
    if (*pStr == '\0'){
        g_nStatus = kValid;
    }
    return (int)num;
}
int StrToInt(char* pStr){
    g_nStatus = kInvalid;
    long long num = 0;
    if (pStr != nullptr || *pStr != '\0'){
        bool minus = false;
        if (*pStr == '+'){
            pStr++;
        }
        else if (*pStr == '-'){
            minus = true;
            pStr++;
        }
        if (*pStr != '\0'){
            num = StrToIntCore(pStr, minus);
        }
    }
    return (int)num;
}

// 和为s的数字:排序数组中找出两个和为s的数字
bool FindNumbersWithSum(int data[], int length, int sum, int* num1, int* num2){
    bool found = false;
    if (data == nullptr || length < 2 || num1 == nullptr || num2 == nullptr){
        return found;
    }
    int ahead = 0;
    int behind = length - 1;
    *num1 = *num2 = 0;
    while (ahead < behind){
        long long currentSum = data[ahead] + data[behind];
        if (currentSum == sum){
            *num1 = data[ahead];
            *num2 = data[behind];
            found = true;
            break;
        }
        else if (currentSum < sum)
            ahead++;
        else
            behind--;
    }
    return found;
}

// 左旋转字符串
void Reverse(char* pBegin, char* pEnd){
    if (pStr == nullptr || pBegin == nullptr || pEnd == nullptr)
        return;
    while (pBegin < pEnd){
        char temp = *pBegin;
        *pBegin = *pEnd;
        *pEnd = temp;
        pBegin++; pEnd--;
    }
}
char* LeftReverseString(char* pStr, int n){
    if (pStr != nullptr){
        int nLength = static_cast<int>(strlen(pStr));
        if (nLength > 0 && n > 0 && n < nLength){
            char* pBegin = pStr;
            char* pEnd = pStr + nLength - 1;
            Reverse(pStr, pBegin, pEnd);
            Reverse(pStr, pBegin, pStr + n - 1);
            Reverse(pStr, pStr + n, pEnd);
        }
    }
    return pStr;
}