[LeetCode]--20. Valid Parentheses-白红宇

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.

public boolean isValid(String s) {        if (s.length() % 2 != 0)            return false;        char[] characters = new char[s.length()];        int index = 0, i;        for (i = 0; i < s.length(); i++) {            if (s.charAt(i) == '(')                characters[index++] = s.charAt(i);            if (s.charAt(i) == '[')                characters[index++] = s.charAt(i);            if (s.charAt(i) == '{') {                characters[index++] = s.charAt(i);                System.out.println(index);            }            if (s.charAt(i) == ')') {                if (--index < 0 || characters[index] != '(')                    break;            }            if (s.charAt(i) == ']') {                if (--index < 0 || characters[index] != '[')                    break;            }            if (s.charAt(i) == '}') {                if (--index < 0 || characters[index] != '{')                    break;            }        }        if (i == s.length() && index == 0)            return true;        return false;    }

方法通过了，不过觉得有点笨，暂时也没想到好的，先就这样啦。LeedCode也没给我们提供详解。

找到一个写得比较好的程序，极力推荐大家不要看我的，看这个，又感觉被甩了几条街。

public boolean isValidParentheses(String s) {        Stack
     
       stack = new Stack
      
       ();        for (Character c : s.toCharArray()) {        if ("({[".contains(String.valueOf(c))) {                stack.push(c);            } else {               if (!stack.isEmpty() && is_valid(stack.peek(), c)) {                   stack.pop();               } else {                   return false;               }           }       }       return stack.isEmpty();    }    private boolean is_valid(char c1, char c2) {        return (c1 == '(' && c2 == ')') || (c1 == '{' && c2 == '}')            || (c1 == '[' && c2 == ']');    }

用了系统的栈，我用的数组栈。