1. 括號匹配

• 堆疊版：

  class Solution(object):
      def isValid(self, s):
          """
          :type s: str
          :rtype: bool
          """
          stack = []
          para_map = {')':'(', ']':'[', '}':'{'}
          for c in s:
              if c not in para_map:
                  stack.append(c)
              elif not stack or stack.pop() != para_map.get(c):
                  return False
          return not stack
   
  

• 字串處理版：

  class Solution(object):
      def isValid(self, s):
          """
          :type s: str
          :rtype: bool
          """
          length = len(s)
          s = s.replace('()', '').replace('[]', '').replace('{}', '')
          while len(s) != length:
              length = len(s)
              s = s.replace('()', '').replace('[]', '').replace('{}', '')
          return length == 0
   
  

  使用 Java 的 do while 語句，形式上將更為簡潔：

  class Solution {
      public boolean isValid(String s) {
          int length;
          do {
              length = s.length();
              s = s.replace("()", "").replace("[]", "").replace("{}", "");
          } while (s.length() != length);
          return length == 0;
      }
  }