C语言匹配字符串中所有小括号

For each string in the array:
    Find the first '{'. If there is none, leave that string alone.
    Init a counter to 0.
    For each character in the string: 
        If you see a '{', increment the counter.
        If you see a '}', decrement the counter.
        If the counter reaches 0, break.
    Here, if your counter is not 0, you have invalid input (unbalanced brackets)
    If it is, then take the string from the first '{' up to the '}' that put the
     counter at 0, and that is a new element in your array.

对于单个的一个字符串来说,从左边向右扫描.对于扫描到的每个'('(记录为p):
    初始化计数器为0.
    对于从p之后的字符进行扫描:
        如果遇到'(',计数器加一.
        如果遇到')',计数器减一.
        如果计数器达到0,退出循环(记录这个字符为q).
    如果循环结束都没有使计数器达到0,则小括号匹配不成对,退出程序.
    从p到q即一段合适的字符串,打印后进入下一个循环.(继续向右扫'(')

/*
 * author:@hit9
 * github:@hit9
 * what-is:find all brackets, and print their postions
 */
#include <stdio.h>

int main(int argc, const char *argv[])
{
    char str[100] = "a(bc)(de(fg)h(jk)s)"; //input string
    char *p = str, *q = NULL, *t = NULL;  //pointer p scan the str

    int count = 0;

    for (; *p != '\0'; p++){
        if (*p  != '(') continue;
        //for each '(' :
        for (q = p; *q != '\0'; q++){
            if (*q == '(') count++; //if see '(', increment count
            if (*q == ')') count--; //if see ')' decrement count
            if (count == 0) break;
        }
        if (count != 0){//unbalanced brackets input
            printf("unbalanced brackets at : %s\n", p);
            return 0;
        }
        for (t = p; t != q+1; printf("%c", *t++));//print characters from '(' to ')'
        printf("\n");
    }
    return 0;
}