{
if (n < 2 || result == NULL)
return 0;
vector f;
f.push_back (0);
f.push_back (1);
for (int i = 2; i <= n; i++)
{
f.push_back(f[i-1]+f[i-2]);
if (f[i] <= 0) //check
return 0;
}
*result = f[n];
return 1;
}
2.给字符串str,将其倒序。例如“abcdef”变为“fedcba”,要求空间复杂度为O(1).
void turn (char * str)
{
if (str == NULL)
return;
int len = strlen (str);
int midd = len / 2;
char tmp;
for (int i = 0; i < midd; i++)
{
tmp = str[i];
str[i] = str[len - 1];