有效的括号

给定一个只包括 ‘(’，’)’，’{’，’}’，’[’，’]’ 的字符串，判断字符串是否有效。有效字符串需满足：
左括号必须用相同类型的右括号闭合。左括号必须以正确的顺序闭合。注意空字符串可被认为是有效字符串。
示例 1:
输入: “()”
输出: true
示例 2:
输入: “()[]{}”
输出: true
示例 3:
输入: “(]”
输出: false
示例 4:
输入: “([)]”
输出: false
示例 5:
输入: “{[]}”
输出: true
1.Java栈

class Solution {

  // Hash table that takes care of the mappings.
  private HashMap<Character, Character> mappings;

  // Initialize hash map with mappings. This simply makes the code easier to read.
  public Solution() {
    this.mappings = new HashMap<Character, Character>();
    this.mappings.put(')', '(');
    this.mappings.put('}', '{');
    this.mappings.put(']', '[');
  }

  public boolean isValid(String s) {

    // Initialize a stack to be used in the algorithm.
    Stack<Character> stack = new Stack<Character>();

    for (int i = 0; i < s.length(); i++) {
      char c = s.charAt(i);

      // If the current character is a closing bracket.
      if (this.mappings.containsKey(c)) {

        // Get the top element of the stack. If the stack is empty, set a dummy value of '#'
        char topElement = stack.empty() ? '#' : stack.pop();

        // If the mapping for this bracket doesn't match the stack's top element, return false.
        if (topElement != this.mappings.get(c)) {
          return false;
        }
      } else {
        // If it was an opening bracket, push to the stack.
        stack.push(c);
      }
    }

    // If the stack still contains elements, then it is an invalid expression.
    return stack.isEmpty();
  }
}

2.C#将括弧依次加入到stack中，如果栈顶的字符和将要加的括弧匹配，那么删除将栈顶字符删除。如果栈中的个数为零，那么为有效；否则，无效。

public class Solution {
    public bool IsValid(string s) {
            if (s==null||s=="")
            {
                return true;
            }
            
            Stack<char> kuohu = new Stack<char>();
            kuohu.Push(s[0]);
            for (int i = 1; i <s.Length; i++)
            {
                if (kuohu.Count==0)
                {
                    kuohu.Push(s[i]);
                    
                }
                else
                {
                    if (OppositeChar(kuohu.Peek()) == ' ')
                    {
                        return false;
                    }
                    if (OppositeChar(kuohu.Peek()) == s[i])
                    {
                        kuohu.Pop();
                    }
                    else
                    {
                        kuohu.Push(s[i]);
                    }
                }
            }
            if (kuohu.Count==0)
            {
                return true;
            }
            return false;
    }
    public  char OppositeChar(char inputChar)
        {
            if (inputChar=='(')
            {
                return ')';
            }
            else if (inputChar=='{')
            {
                return '}';
            }
            else if (inputChar=='[')
            {
                return ']';
            }
            return ' ';
        }
}