/*
* Exercise 2-5 Page 48
*
* Write the function any(s1,s2), which returns the first location
* in the string s1 where any character from the string s2 occurs,
* or -1 if s1 contains no characters from s2. (The standard library
* function strpbrk does the same job but returns a pointer to the
* location.)
*
*/
Here's a much better solution, by Partha Seetala. This solution has a worst-
case time complexity of only O(n + m) which is considerably better.
1：类似junit的断言，只是在assert中的断言，如果不满足的话就程序退出。
比如
#include <assert.h>
int main(void)
{
assert(6 < 5);
system("pause");
return 0;
}
在执行到assert(6 < 5);
的时候因为不满足断言，于是程序退出。
如果不想让assert(6 < 5)起作用，就在最上面添加宏定义# ......
