20. Valid Parentheses

解题思路：

栈思想，从左往右遍历，如果是左括号则入栈，右括号则出栈，最后判断栈是否为空，空则为有效括号组，否则无效。

实现代码：

// 20. Valid Parentheses
bool isValid(string s) {
  stack<char> brackets;
  map<char, char> bracketMap;
  bracketMap[')'] = '(';
  bracketMap[']'] = '[';
  bracketMap['}'] = '{';
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(' || s[i] == '[' || s[i] == '{') {
      brackets.push(s[i]);
    } else if (s[i] == ')' || s[i] == ']' || s[i] == '}') {
      if (!brackets.empty() && brackets.top() == bracketMap[s[i]]) {
        brackets.pop();
      } else {
        return false;
      }
    }
  }
  if (brackets.empty()) return true;
  else return false;
}

问题描述：

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

An input string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. 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