Stack push-in

Enter Two integer sequences. the first sequence indicates the order in which the stack is pushed. Determine whether the second sequence is in the pop-up order of the stack. Assume that all numbers pushed into the stack are not equal. For example, the sequence 1, 2, 3, 4, and 5 is the order in which a stack is pushed. The sequence 4, 5, 3, 2, and 1 is a pop-up sequence corresponding to the stack sequence, but the sequence 4, 3, 5, 2. It cannot be the pop-up sequence of the Pressure stack sequence. Input: each test case contains three rows: the first row is an integer n (1 <= n <= 100000), indicating the length of the sequence. The second row contains n integers, indicating the order of stack pushing. The third row contains n integers, indicating the pop-up sequence of the stack. Output: corresponds to each test case. If the second sequence is the pop-up sequence of the first sequence, output Yes; otherwise, output No. Sample input: 51 2 3 4 54 5 3 2 151 2 3 4 54 3 5 1 2 sample output: YesNo code AC: idea: scan + inbound stack + Stack: the current output is equal to the top element of the stack. If the current output is not the top element of the stack, it is necessary to import the element into the stack until this element is generated. If this element is not found in the stack, this is the sequence of errors. [Cpp] # include <stdio. h> # include <stdlib. h> int main () {int n, * stack1, * stack2, * stack, I, j, top =-1, flag, f; while (scanf ("% d", & n )! = EOF) {stack1 = (int *) malloc (sizeof (int) * n); stack2 = (int *) malloc (sizeof (int) * n ); stack = (int *) malloc (sizeof (int) * n); for (I = 0; I <n; I ++) {scanf ("% d ", & stack1 [I]) ;}for (I = 0; I <n; I ++) {scanf ("% d", & stack2 [I]);} top =-1; I = 0; j = 0; flag = 1; while (stack1 [I]! = Stack2 [j]) {stack [++ top] = stack1 [I]; if (++ I) >=n) {printf ("No \ n "); flag = 0; break;} if (! Flag) {continue;} I ++; j ++; while (j <n) {f = 0; if (top>-1) {if (stack2 [j] = stack [top]) {top --; j ++;} else {f = 1 ;}} else {f = 1 ;} if (f) {while (stack2 [j]! = Stack1 [I]) {stack [++ top] = stack1 [I]; I ++; if (I> = n) {printf ("No \ n "); flag = 0; break; }}if (flag) {stack [++ top] = stack1 [I] ;}} if (! Flag) {break ;}} if (! Flag) {continue;} printf ("Yes \ n");} return 0 ;}