implementation

I have already seen two stack implementation of this question but I am really confused as how one could get O(1) operation. consider following example: S1[3542761986759] S2[3332221111111] The idea/algorithm here is Push element E on S1 Check to see if top of S2 >= E and if true, insert E on S2 But when getMin is called, we return top of S2 but that leaves S2 in weird state as S2 contains repeated current min elements, so the solution is to search next minimum element in S2 and return it. This is O(n). Can anyone please help me to understand the solution? Using a linked list store the current

I'm not sure if its the right title but ill explain what i mean. I'm making more than one android application, but they have the same structure sliding menu , list view , about me , costume dialog with (copy,share,like) with some modifications in the styles (colors, backgrounds , fonts , menu strings ) my qustion is : is there any way to use the structure as library or implemntation or anu other way insted of coping the same codes in every projects ? Basically what you want to do is to create an Android library . Just develop it like a normal application, with Activities and layout . Please

In an application I am developing I use splice on Linux for socket-to-socket data transfer. Do other operating systems (specifically at least Windows, OS X and FreeBSD) implement splice or an equivalent solution? Is it possible to imitate socket-to-socket data splice ing on Windows with sendfile ¹ + memmap ¹? ¹ Both exist on Windows under different names which I do not remember. Update You can see the performance improvements of splice vs user space buffers on Linux. DF , DR , F , MF , MR are my application in its different tunneling modes, NX is NGINX web server -p+t uses the Linux system

Does anyone have an idea how is the method/function Int() or floor() implemented? What I am looking for a respective implementation as the following is for abs() function. Int Abs (float x){ if x > 0 return x; else return -x } I am struggling to find a solution for it without using the modulus operator. TheJubilex Seems to me like floor(n) = n - (n % 1) should do the trick. Using the IEEE 754 binary floating point representation one possible solution is: float myFloor(float x) { if (x == 0.0) return 0; union { float input; // assumes sizeof(float) == sizeof(int) int output; } data; data.input