ACM Brace pairing problem

Input: The first line enters a number n (0<n<=100), indicating that there are n sets of test data. The next n lines enter multiple sets of input data, each set of input data is a string s (S is less than 10000, and S is not an empty string), and the number of test data groups is less than 5 groups. The data guarantees that s contains only "[", "]", "(", ")" four characters.

Output: The output of each set of input data is a row, if the string contains parentheses are paired, the output is yes, if not paired output no

Sample input:

3[(]) (]) ([[[] ()])
Sample output:

Nonoyes

This is the Nanyang Polytechnic Institute of ACM Exercises, my procedures are as follows:
In practicing this problem, I learned the knowledge and usage of ① stack; the method of opening and releasing dynamic space of② two-dimensional array;③ The most important thing is to learn this way of thinking, Mark.

#include <stdio.h>#include<stdlib.h>#include<string.h>#defineTRUE 1#defineFALSE 0#defineStackSize 10000typedefstruct{    CharData[stacksize]; inttop;} Sqstack;intNUM1 =0, num2 =0;voidInitstack (Sqstack *s);intStackempty (Sqstack *s);intStackfull (Sqstack *s);intPush (Sqstack *s,Charx);CharPop (Sqstack *s);intStacktop (Sqstack *s,Char*x);voidBracket (sqstack *s,Char*a);intMainvoid){    intN =0; inti =0, j =0, k =0; int*length; Char**s;        Sqstack S; scanf ("%d",&N); //Open two-bit array dynamic memory space, n rows 10000 columnss = (Char**)malloc(sizeof(Char*)*N);  for(i =0; I < n;i++) S[i]= (Char*)malloc(sizeof(Char)*10000);  for(i =0; I < n;i++)    {        if('\ n'==GetChar ()) {scanf ("%s", S[i]); }    }     for(i =0; I < n;i++) {bracket (&S, s[i]); if(NUM1 = =0&& num2 = =0) {printf ("yes\n"); }        Elseprintf ("no\n"); NUM1= Num2 =0; } System ("Pause");  for(i =0; I < n;i++)         Free(S[i]);  Free(s); return 0;}voidBracket (sqstack *s,Char*a) {    intI,j,length =strlen (a);    Initstack (s);  for(i =0; I < length;i++)    {        Switch(A[i]) { Case '[':P Ush (S,a[i]); num1++; Break;  Case '(':P Ush (S,a[i]); num2++; Break;  Case ']':                if('['= = Pop (s) | | FALSE = =Pop (s)) {NUM1--; }                 Break;  Case ')':                if('('= = Pop (s) | | FALSE = =Pop (s)) {num2--; }                 Break; default: Break; }    }}voidInitstack (Sqstack *s) {s->top =-1;//Initialize the order stack to be empty}intStackempty (Sqstack *R) {    if(S->top = =-1)        returnTRUE; Else        returnFALSE;}intStackfull (Sqstack *s) {    if(S->top = = stacksize-1)        returnTRUE; Else        returnFALSE;}intPush (Sqstack *s,Charx) {    if(Stackfull (s)) {//printf ("stackfull~\n");        returnFALSE; }    Else{s->top++; S->data[s->top] =x; returnTRUE; }}CharPop (Sqstack *s) {    if(Stackempty (s)) {//printf ("stackempty~\n");        returnFALSE; }    Else    {        returns->data[s->top--]; }}intStacktop (Sqstack *s,Char*x) {    if(Stackempty (s)) {//printf ("stackempty~\n");        returnFALSE; }    Else    {        *x = s->data[s->top]; returnTRUE; }}

ACM Brace pairing problem