题解 | #字符串变形#

/**
 * 
 * @param s string字符串 
 * @param n int整型 
 * @return string字符串
 */
void convertion(char *s, int n)   //将字符串的大小写进行转换
{
    int i = 0;
    int j = 0;
    for (i = 0, j = n - 1; i < j; i++, j--)
    {
        if ((s[i] <= 'Z') && (s[i] >= 'A'))
        {
            s[i] += 32;
        }
        else if ((s[i] <= 'z') && (s[i] >= 'a'))
        {
            s[i] -= 32;
        }
        if ((s[j] <= 'z') && (s[j] >= 'a'))
        {
            s[j] -= 32;
        }
        else if ((s[j] <= 'Z') && (s[j] >= 'A'))
        {
            s[j] += 32;
        }
    }
    if (n % 2 != 0)
    {
        if ((s[i] <= 'Z') && (s[i] >= 'A'))
        {
            s[i] += 32;
        }
        else if ((s[i] <= 'z') && (s[i] >= 'a'))
        {
            s[i] -= 32;
        }
    }
}

char* trans(char* s, int n ) {
    // write code here
    char a[10000000] = {0};
    int j = 0;
    convertion(s, n);       //将字符串的大小写进行转换
    for (int i = n - 1; i >= 0; i--)
    {
        if (s[i] == ' ')
        {
            strcpy(&a[j], &s[i+1]);
            a[n-i-1] = ' ';
            j = n-i;
        }
        if (i == 0)
        {
            strcpy(&a[j], &s[i]);
        }
    }
    a[n] = '\0';
    strcpy(s, a);
    return s;
}