Data structure and algorithm stack and queue two: Stack push-in, pop-up sequence

At that time I was learning this is also very do not understand this problem, a stack of indentation and pop-up sequence of the judgment of a look not to know, but also to determine what to do. As long as the last-in-first-out rule is met. But here's what I'm going to say briefly about this press-in and pop-up sequence. Is the two sequence we have given the hypothesis, one for the press-in sequence and one for the pop-up sequence. Then we go through an auxiliary stack, pushing the data into the sequence into a temporary secondary stack, if the top element of the stack is just like the data from the pop-up sequence, then we pop up, and if not, we'll press the data in the sequence into the temporary stack until we get to the end of the sequence. If the press-in sequence ends, all data pops up on the temporary stack, which is a popup press-in or pop-up sequence. Directly understand:

The code is as follows:

 Public BooleanIspoporder (intArr1[],intarr2[]) {    //determine if two sequences have an empty sequence    if(Arr1.length = = 0 | | arr2.length = = 0){          return false; }    //Create a temporary secondary stackstack<integer> s =NewStack<>(); //press the data into the stack and compare it     for(inti = 0,j = 0;i <arr1.length;) {S.push (arr1[i]); //determines whether the top element of the stack matches the pop-up sequence          while(S.size () > 0 && s.peek () = =Arr2[j])              {S.pop (); J++; }    }    returns.isempty ();}

The code above me uses an array to implement the sequence. ARR1 represents a press-in sequence, and arr2 represents a pop-up sequence. The key thing we need to catch is the pop-up comparison on the loop at the press-in, with the help of a temporary stack to implement the algorithm.

Data structure and algorithm stack and queue two: Stack push-in, pop-up sequence