LeetCode集合的妙用！

一、LeetCode-20、有效的括号

• 解题思路：这里充分利用了容器，首先，将成对的括号用map存储起来，且key存储闭括号，方便使用contains（）方法，其中前半括号都存储到栈里面，遇到闭括号就从栈种取出来一个与该闭括号的value指进行比较看看是否相等，不等 或者最后栈不为空则说明该字符串不合格。

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();
  }
}

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/valid-parentheses/solution/you-xiao-de-gua-hao-by-leetcode/