括号配对问题

描述：

　　　现在，有一行括号序列，请你检查这行括号是否配对。

输入:

第一行输入一个数N（0<N<=100）,表示有N组测试数据。后面的N行输入多组输入数据，每组输入数据都是一个字符串S(S的长度小于10000，且S不是空串），测试数据组数少于5组。数据保证S中只含有"[","]","(",")"四种字符

输出:

每组输入数据的输出占一行，如果该字符串中所含的括号是配对的，则输出Yes,如果不配对则输出No

样例输入：

3[(])(])([[]()])

样例输出：

NoNoYes

---

import java.util.Scanner;public class No2 {    PRivate String checkout(String str) {        try {            StringBuffer asb = new StringBuffer(str);  // 模拟栈A            StringBuffer bsb = new StringBuffer();   // 模拟栈B            int length = asb.length();            while (length > 0) {                // 遇右括号，移A入B                if (asb.charAt(length - 1) == ')'                        || asb.charAt(length - 1) == ']') {                    bsb.append(asb.charAt(length - 1));                    asb.deleteCharAt(length - 1);                    length = asb.length();                    continue;                }                // 遇左括号，若无法与B中栈顶元素合法匹配，则序列非法；若合法匹配，移A移B                if (asb.charAt(length - 1) == '(') {                    if (bsb.charAt(bsb.length() - 1) != ')') {                        return "No";                    } else {                        asb.deleteCharAt(asb.length() - 1);                        bsb.deleteCharAt(bsb.length() - 1);                        length = asb.length();                        continue;                    }                }                if (asb.charAt(length - 1) == '[') {                    if (bsb.charAt(bsb.length() - 1) != ']') {                        return "No";                    } else {                        asb.deleteCharAt(asb.length() - 1);                        bsb.deleteCharAt(bsb.length() - 1);                        length = asb.length();                        continue;                    }                }            }            if (bsb.length() == 0) {                return "Yes";            }        } catch (RuntimeException e) {        }        return "No";    }    public static void main(String args[]) throws Exception {        try {            Scanner cin = new Scanner(System.in);            int num = cin.nextInt();            String[] result = new String[num];            No2 no2 = new No2();            for (int i = 0; i < num; i++) {                String str = cin.next();                result[i] = no2.checkout(str);            }            for (int i = 0; i < num; i++) {                System.out.println(result[i]);            }        } catch (RuntimeException e) {        }    }}