PHP implementation of Stack press and pop-up sequence

This article mainly introduces the PHP implementation of the stack and pop-up sequence, has a certain reference value, now share to everyone, the need for friends can refer to

Title Description

Enter a sequence of two integers, and the first sequence represents the stacking order of the stack, judging whether the second sequence is the pop-up order for the stack. Assume that all the numbers that are pressed into the stack are not equal. For example 1，2，3，4，5 , a sequence is the indentation order of a stack, and the sequence 4，5，3，2，1 is a pop-up sequence corresponding to the stack sequence, but 4，3，5，1，2 it cannot be the pop-up sequence of the stack sequence. (Note: The lengths of the two sequences are equal)
Time limit: 1 seconds space limit: 32768K

• Thinking of solving problems

• Passing in the stack sequence, and the stack sequence, we use a stack as a secondary stack, traversing the stack sequence into the secondary stack, if the span of the auxiliary stack is not empty, and the top element of the stack is equal to the current stack element, the auxiliary stack element is ejected. If the secondary stack is empty after the traversal, the second sequence is the pop-up order of the stack

• The code is as follows

<?phpfunction Ispoporder ($pushValue, $popValue) {    $stack = new Splstack;    $count = count ($pushValue);    for ($i = 0, $j = 0; $i < $count; $i + +) {            $stack->push ($pushValue [$i]);        while (! $stack->isempty () && $stack->top () = = $popValue [$j] && $j < $count) {                    $stack Pop ();                    $j + +;        }    }    return $stack->isempty ();} Var_dump (Ispoporder ([1, 2, 3, 4, 5], [4, 5, 3, 2, 1]));

• Instance commentary
  

• Stack sequence for 1,2,3,4,5 out of stack sequence for 4,5,3,2,1

• First traversal, obviously 1 not equal to 4 continue traversal, traverse 4 times to the secondary stack of the stack element is 1,2,3,4 stack top element is 4 when

• Bring up the top element of the auxiliary stack, the stack element will be left to 5 /code>

• At this point the stack element of the secondary stack is 3 on the top of the stack, and obviously 3 is not equal to the stack element 5 , we continue to traverse the 5 into the secondary stack

• Just when the stack element is the same as 5 and the top element of the stack, so we pop up the top element of the stack, and the stack element becomes 3

• To continue, the top element of the stack is now 3 the same as the stack element, pop up the top element of the stack

• This time the top element of the stack becomes 2 , and the stack element becomes 2 /code>, continue to pop the top of the stack

• When the top element of the stack changes to 1 , the stack element becomes 1 , the top element of the stack pops up

• Because the secondary stack is now empty Jump out while

• Because the stack element is at this point to all enter the secondary stack out for

• Secondary stack eventually empty, program end