19. Remove Nth Node From End of List

解题思路：

快慢指针思想，让两个指针 p, q都指向头结点，p 先向后移动 n + 1 步，然后p, q一起向后移动，当 p 到达尾结点时，q指向目标节点的前驱结点，做删除操作，然后按照题目要求返回头结点即可。

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