几个注意的小算法

看C程序设计语言这本书时，感觉大神写的程序逻辑性确实很强。很多小细节值得我们学习。

1.转换十六进制。

int hexalpha_to_int(int c)
{
    char hexalpha[] = "aAbBcCdDeEfF";
    int i;
    int answer = 0;

    for(i = 0; answer == 0 && hexalpha[i] != '\0'; i++)
    {
        if(hexalpha[i] == c)
        {
            answer = 10 + (i / 2);
        }
    }

    return answer;
}

unsigned int htoi(const char s[])
{
    unsigned int answer = 0;
    int i = 0;
    int valid = 1;
    int hexit;

    if(s[i] == '0')
    {
        ++i;
        if(s[i] == 'x' || s[i] == 'X')
        {
            ++i;
        }
    }
    /**
     * 这个地方的算法是这样的：
     * 0X4567换成16进制要进行这样的运算  4*16^3 + 5*16^2 + 6*16^1 + 7*16^0
     * 而这里是这样进行运算的  先记录最高位 answer = 4
     * 然后 (((4*16 + 5) * 16) + 6) * 16 + 7
    */
    while(valid && s[i] != '\0')
    {
        answer = answer * 16;
        if(s[i] >= '0' && s[i] <= '9')
        {
            answer = answer + (s[i] - '0');
        }
        else
        {
            hexit = hexalpha_to_int(s[i]);
            if(hexit == 0)
            {
                valid = 0;
            }
            else
            {
                answer = answer + hexit;
            }
        }

        ++i;
    }

    if(!valid)
    {
        answer = 0;
    }

    return answer;
}

2.编写函数any(s1,s2),将字符串s2中的任意字符在s1中第一次出现的位置返回(关键是时间复杂度 O(m + n)  自己写的时候是O（mn）

int any(char *s1, char *s2) 
{
        char array[256]; /* rjh comments
                          * (a) by making this char array[256] = {0}; the first loop becomes unnecessary.
                          * (b) for full ANSIness, #include <limits.h>, make the array unsigned char,
                          *     cast as required, and specify an array size of UCHAR_MAX + 1.
                          * (c) the return statements' (parentheses) are not required.
                          */
        int  i;
        if (s1 == NULL) {
                if (s2 == NULL) {
                        return(0);
                } else {
                        return(-1);
                }
        }
 
        for(i = 0; i < 256; i++) {
                array[i] = 0;
        }
        
        while(*s2 != '\0') {
                array[*s2] = 1;
                s2++;
        }
 
        i = 0;
        while(s1[i] != '\0') {
                if (array[s1[i]] == 1) {
                        return(i);
                }
                i++;
        }
        return(-1);
}

3.编写一个函数rightrot(x,n),该函数返回将x循环右移（及从右端移出的位从最左端移入）n位返回的值。（充分利用了移位的特性）

unsigned rightrot(unsigned x, unsigned n)
{
    while (n > 0) {
        if ((x & 1) == 1)
            x = (x >> 1) | ~(~0U >> 1);
        else
            x = (x >> 1);
        n--;
    }
    return x;
}