华为的最新的两道算法设计题

要求均用C语言实现

1.找出最大长度子字符串（只包含字母），打印并且返回长度。 例如 str = "abc123abcd234abcdefgha324adsdawqdasdaseqqwe345abchded"，最大子字符串是

“adsdawqdasdaseqqwe”

int findMaxSubstring(char* str)
{
    int maxLength = 0;
    int maxStartIndex = 0;
    int curLength = 0;
    int curStartIndex = 0;
    bool isFind = 0;
    for(unsigned int i = 0;i<strlen(str);i++)
    {
        if(str[i] >= 'a' && str[i] <= 'z')
        {
            if(isFind == 0)
            {
                isFind = 1;
                curLength = 1;
                curStartIndex = i;
            }
            else
            {
                curLength++;
            }
        }
        else if (str[i] < 'a' || str[i] > 'z')
        {
           isFind = 0;
           if(curLength > maxLength)
           {
              maxLength = curLength;
              maxStartIndex = curStartIndex;
              curLength = 0;
           }
        }
    }
    char *p = NULL;
    p = &str[maxStartIndex];
    while(*p >= 'a' && *p <= 'z')
    {
        putchar(*p);
        p++;
    }
    return maxLength;
}

2. 一个四位数，如1024，1004，打印出他们的中文形式，如果一千零二十四，一千零四

 这题因为限定了4位数，所以只考虑了4位数的情况，吃点分享一个大小写转换的源码，里面有不限位数的情况，当时调试的很痛苦，思想差不多。

void iConvert(int digit)
{
    char a[5][10] = {"千","百","十","","零"};
    char b[11][10] = {"零","一","二","三","四","五","六","七","八","九","十"};
    char result[50] = {''};
    int A[4] = {};
    for(int i=3;i>=0;i--)
    {
        A[i] = digit % 10;
        digit = int(digit/10);
    }
    printf("%d,%d,%d,%d
",A[0],A[1],A[2],A[3]);
    int foundZero = 0;
    for(int i = 0 ;i<4;i++)
    {
        if(A[i]>0)
        {
            strcat(result,b[A[i]]);
            strcat(result,a[i]);
        }
        if(A[i]==0 && foundZero == 0)
        {
           if(i!=3)//如果不是最后一位，则不追加零
           {
             strcat(result,a[4]);
             foundZero = 1;
           }
        }    
    }
    puts(result);
}

运行结果：