Given a string containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.

The brackets must close in the correct order, “()” and “()[]{}” are all valid but “(]” and “([)]” are not.

#include <stdio.h>
#include <string.h>
//資料結構：棧
//演算法：抓取到左括號，就將它推入棧(push)；抓取到右括號，就pop，取出一個左括號，判斷是否匹配，如果匹配則繼續；不匹配立即結束迴圈，輸出false。
 

int stack[1000];
int len_stack;

int pop(void)
{
//自己想   
}

void push(int new_element)
{
   //自己想
 }

int trans(char c)//把括號轉化為數字來處理
{
    switch (c)
    {
        case '{':return 1;
        case '}':return 2;
        case '[':return 3;
        case ']':return 4;
        case '(':return 5;
        case ')'
 
:return 6;
    }

}

int main()
{
    char exp[10000];
    scanf("%s",&exp);
    long len_exp =strlen(exp);

    int boo=1;//boo為1代表匹配，為0代表不匹配

    int t,i;

    len_stack=0;

    for (i=0;i<len_exp;i++)
    {
        t=trans(exp[i]);
        if (t%2==0) 
        {
            if (len_stack==0) //如果棧中沒了元素，說明右括號前面沒有左括號，  
 

            {
                boo=0;
                break;
            }

            else
            {
                if (pop()+1!=t) 
                {
                    boo=0;
                    break;
                }

            }
        }
        else  push(t);

    }
    if (len_stack!=0) boo=0;

    if (boo==0) printf("False\n");
        else    printf("True\n");
}