Valid Parentheses

有效的括号

题目

Given a string containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[‘ and ‘]’, determine if the input string is valid.

An input string is valid if:

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.

Note that an empty string is also considered valid.

Example 1:

Input: “()”
Output: true

Example 2:

Input: “()[]{}”
Output: true

Example 3:

Input: “(]”
Output: false

Example 4:

Input: “([)]”
Output: false

Example 5:

Input: “{[]}”
Output: true

解析重点

1.这道题最重要是找出括号之间的规律，当有左括号的时候不用管，出现右括号的时候要检查左边是否有对应的括号，
有的话将其消除掉，不断重复这个过程。但是，这需要我们从内到外处理数据，我们可以借助栈来处理这种情况。
2.循环处理表达式的每个括号，当遇到左括号时，我们将其推入栈。
3.当遇到右括号时，弹出栈顶元素，看是否匹配，如果无法消除，则表达式无效，如果匹配，则消除这对数据。
4.如果最后站内任有元素，表达式无效。

java代码

具体java代码参考官方

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();
  }
}