记腾讯面试题-整形压缩

2020-05-25  本文已影响0人  布灵不灵的丙丙

背景:

实现方式:

编码实现:

void chartobs(char  ch, char* ps)
{
    const int  size = 8 * sizeof(char);
    for (int i = 7; i >= 0; i--, ch >>= 1)
    {
        ps[i] = (01 & ch) + '0';
    }
    ps[size] = '\0';
}

void itobs(int  n, char* ps)
{
    const int  size = 8 * sizeof(int);
    for (int i = size - 1; i >= 0; i--, n >>= 1)
    {
        ps[i] = (01 & n) + '0';
    }
    ps[size] = '\0';
}

void show_bstr(const char* str)
{
    size_t i = 0;
    while (str[i])
    {
        putchar(str[i]);
        if (++i % 8 == 0 && str[i])
            putchar(' ');
    }
    printf("\n");
}
//将一个四字节的整形数值压缩成1~5个字节
size_t encode(unsigned int num, char *buf)
{
    size_t len = 0;
for (int a = sizeof(unsigned int); a >= 0; a--)
    {
        char c;
        c = num >> (a * 7) & 0x7f;
        if (c == 0x00 && len == 0)
        {
            char szDest[16] = { 0 };
            chartobs(c, szDest);
            cout << szDest << " ";
            continue;
        }
        //处理高位标识
        if (a == 0)
            c &= 0x7f;
        else
            c |= 0x80;
        buf[len] = c;
        len++;

        char szDest[16] = { 0 };
        chartobs(c, szDest);
        cout << szDest << " ";
    }
    cout << endl;
    if (len == 0)
    {
        len++;
        buf[0] = 0;
    }
    return len;
}

//将一个1~5个字节的值还原成四字节的整形值
unsigned int decode(char *buf, size_t len)
{
    unsigned int num = 0;
    for (int index = 0; index < (int)len; index++)
    {
        char c = *(buf + index);
        num = num << 7;

        c &= 0x7f;
        num |= c;
    }
    cout << "decode: " << num << endl;
    return num;
}

测试用例:

int main(void)
{
    unsigned int num = 300;
    char szShowStr[64] = { 0 };
    itobs(num, szShowStr);
    show_bstr(szShowStr);

    char szDest[5] = { 0 };
    size_t len = encode(num, szDest);

    system("pause");
    return 0;
}

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