[编程题]括号匹配问题

关于括号匹配问题可以扩展出来好几个问题：

• 传统的括号匹配，给出一个字符串，判断里面的括号是否匹配
• 给你一个字符串，里面只包含”(“,”)”,”[“,”]”四种符号，请问你需要至少添加多少个括号才能使这些括号匹配起来。

1 传统的括号匹配

给定字符串，输出括号是否匹配，例如：

)
false
()
true
(a)
true
[]{{}}(aa)
true

解决这个问题最经典的就是用栈来实现

public class Test {    public static void main(String[] args){        Scanner scanner = new Scanner(System.in);        String str = "";        while(scanner.hasNext()){            str = scanner.nextLine();            if ("Q".equals(str)) {                break;            }            System.out.println(match(str));        }    }    private static boolean match(String str) {        Stack<Character> stack = new Stack<>();        char[] chars = str.toCharArray();        for (int i = 0; i < chars.length; i++) {            try {                switch (chars[i]) {                case '(':                    stack.add('(');                    break;                case '[':                    stack.add('[');                    break;                case '{':                    stack.add('{');                    break;                case ')':                    if (stack.peek() == '(') {                        stack.pop();                    }                    break;                case ']':                    if (stack.peek() == '[') {                        stack.pop();                    }                    break;                case '}':                    if (stack.peek() == '{') {                        stack.pop();                    }                    break;                default:                    break;                }            } catch (EmptyStackException e) {                return false;            }        }        return stack.empty();    }}

2

这道题转载自：http://blog.csdn.net/beiyeqingteng/article/details/7695274
给你一个字符串，里面只包含”(“,”)”,”[“,”]”四种符号，请问你需要至少添加多少个括号才能使这些括号匹配起来。
如：
[]是匹配的，所需括号个数为 0.
([])[]是匹配的， 所需括号个数为 0.
((]是不匹配的， 所需最少括号个数为 3.
([)]是不匹配的，所需最少括号个数为 2.

public static void minBrace(String s) {      int size = s.length();      // we begin with mb[1][1]      int[][] mb = new int[size + 1][size + 1];      for(int i = 1; i <= size; ++i){            mb[i][i] = 1;        }      // d refers to the distance between i and j, that is d = j - i      for(int d = 1; d < size; d++){            for (int i = 1; i + d <= size; i++) {              int j = i + d;              // the worst case              mb[i][j] = Math.min(mb[i][j - 1], mb[i + 1][j]) + 1;              // the case in which a char between i and j (= i + d) matches              // the character at position j + 1              for (int k = i ; k <= j - 1; k++ ) {                  if (match(s.charAt(k - 1), s.charAt(j - 1)) == true) {                                        mb[i][j] = Math.min(mb[i][j], mb[i][k - 1] + mb[k + 1] [j - 1]);                  }               }          }      }      System.out.println(mb[1][size]);    } static boolean match(char a, char b){        if(a == '(' && b == ')')            return true;        if(a == '[' && b == ']')            return true;        return false;    }