LeetCode 20 - Valid Parentheses | Darren's Blog

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) {

        Stack<Character> stack = new Stack<Character>();

        int n = s.length();

        for (int i = 0; i < n; i++) {

            char c = s.charAt(i);

            if (c == '(' || c == '[' || c == '{')   // Opening parenthesis; always put in the stack

                stack.push(c);

            else {          // Closing parenthesis

                if (stack.isEmpty())    // No more opening parenthesis to match the closing one

                    return false;

                char top = stack.peek();

                if ((top == '(' && c != ')') || (top == '[' && c != ']') || (top == '{' && c != '}'))

                    // The parenthesis at the top of the stack does not match the closing one

                    return false;

                stack.pop();        // A match occurs; get the opening parenthesis out of the stack

            }

        }

        if (stack.isEmpty())    // no opening parenthesis left unmatched

            return true;

        else

            return false;

    }

https://wxx5433.gitbooks.io/interview-preparation/content/part_ii_leetcode_lintcode/stack/valid_parenthese.html

We can use a stack to push the left parentheses into stack and pop the top element from the stack if the new character is right parentheses.

There is a small optimization. If the element in the stack is more than the number of left characters, we can directly return false, since it is impossible to be valid.

Corner Case

1. more right parenthese than left parentheses, stack may become empty and poping elemenet can cause exception.
2. less right parenthese, at last, stack may not be empty

  public boolean isValidParentheses(String s) {
    Stack<Character> stack = new Stack<>();
    for (char c : s.toCharArray()) {
      if (c == '(' || c == '[' || c == '{') {
        stack.push(c);
      } else if (stack.isEmpty() || !isMatch(stack.pop(), c)) {
        return false;
      }
    }
    return stack.isEmpty();
  }

  private boolean isMatch(char c1, char c2) {
    if ((c1 == '(' && c2 == ')') || (c1 == '[' && c2 == ']') 
        || (c1 == '{' && c2 == '}')) {
      return true;
    } 
    return false;
  }

https://discuss.leetcode.com/topic/27572/short-java-solution

public boolean isValid(String s) {
 Stack<Character> stack = new Stack<Character>();
 for (char c : s.toCharArray()) {
  if (c == '(')
   stack.push(')');
  else if (c == '{')
   stack.push('}');
  else if (c == '[')
   stack.push(']');
  else if (stack.isEmpty() || stack.pop() != c)
   return false;
 }
 return stack.isEmpty();
}

https://discuss.leetcode.com/topic/7813/my-easy-to-understand-java-solution-with-one-stack

you need !stack.empty() to handle invalid input and because you can't peek when the stack is empty.

    public boolean isValid(String s) {
        Stack<Character> stack = new Stack<Character>();
        // Iterate through string until empty
        for(int i = 0; i<s.length(); i++) {
            // Push any open parentheses onto stack
            if(s.charAt(i) == '(' || s.charAt(i) == '[' || s.charAt(i) == '{')
                stack.push(s.charAt(i));
            // Check stack for corresponding closing parentheses, false if not valid
            else if(s.charAt(i) == ')' && !stack.empty() && stack.peek() == '(')
                stack.pop();
            else if(s.charAt(i) == ']' && !stack.empty() && stack.peek() == '[')
                stack.pop();
            else if(s.charAt(i) == '}' && !stack.empty() && stack.peek() == '{')
                stack.pop();
            else
                return false;
        }
        // return true if no open parentheses left in stack
        return stack.empty();
    }

https://wxx5433.gitbooks.io/interview-preparation/content/part_ii_leetcode_lintcode/stack/valid_parenthese.html

The string can contains other characters in addition to braces. For example "if main() {}" is valid.

In addition, there are many different types of braces.

Since there are many different types of braces, it is important to make the code easy to change. We can put all possible mapping of braces in a CSV file and read it into a map.

  private static final Map<Character, Character> leftToRightBraces = new HashMap<>();
  private static final Set<Character> leftBracesSet = new HashSet<>();
  private static final Set<Character> rightBracesSet = new HashSet<>();

  static {
    // It is possible to change this to read from, for example a CSV file, to get 
    // more kinds of braces. The benefit of this is that you don't need to change 
    // other code. Simply changes the CSV file, and anything else remains the same. 
    leftToRightBraces.put('(', ')');
    leftToRightBraces.put('[', ']');
    leftToRightBraces.put('{', '}');
    for (char c : new char[] {'(', '[', '{'}) {
      leftBracesSet.add(c);
    }
    for (char c : new char[] {')', ']', '}'}) {
      rightBracesSet.add(c);
    }
  }

  public boolean isValidParentheses(String s) {
    Stack<Character> stack = new Stack<>();
    for (char c : s.toCharArray()) {
      if (isLeftBrace(c)) {
        stack.push(c);
      } else if (isRightBrace(c)) {
        if (stack.isEmpty() || !isMatch(stack.pop(), c)) {
          return false;
        }
      }
    }
    return stack.isEmpty();
  }

  private boolean isLeftBrace(char c) {
    return leftBracesSet.contains(c);
  }

  private boolean isRightBrace(char c) {
    return rightBracesSet.contains(c);
  }

  private boolean isMatch(char c1, char c2) {
    if (leftToRightBraces.containsKey(c1)) {
      return leftToRightBraces.get(c1) == c2;
    }
    return false;
  }

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