一.解法

https://leetcode-cn.com/problems/valid-parentheses/
要点：栈

Python，C++，Java用的方法相同，遍历字符串，对每个字符如果是左括号直接加入栈，如果是右括号则要取栈顶元素比较是否是匹配括号

二.Python实现

class Solution:
    def isValid(self, s: str) -> bool:
        stack = []

        hashmap = {")": "(", "}": "{", "]": "["}

        for char in s:

            if char in hashmap:
                top_element = stack.pop() if stack else '#'
                if hashmap[char] != top_element:
                    return False
            else:
                stack.append(char)

        return not stack

三.C++实现

class Solution {
public:
    bool isValid(string s) {
        
        stack<char> a;
        if(s.length()==0) return true;
        
        for(int i=0;i<s.length();i++){
            if(a.empty()) a.push(s[i]);
            else if(a.top()=='('&&s[i]==')') a.pop();
            else if(a.top()=='['&&s[i]==']') a.pop();
            else if(a.top()=='{'&&s[i]=='}') a.pop();
            else a.push(s[i]);
        }
          
        return a.empty();
        
    }
};

四.java实现

class Solution {

  private HashMap<Character, Character> mappings;

  public Solution() {
    this.mappings = new HashMap<Character, Character>();
    this.mappings.put(')', '(');
    this.mappings.put('}', '{');
    this.mappings.put(']', '[');
  }

  public boolean isValid(String s) {

    Stack<Character> stack = new Stack<Character>();

    for (int i = 0; i < s.length(); i++) {
      char c = s.charAt(i);

      if (this.mappings.containsKey(c)) {

        char topElement = stack.empty() ? '#' : stack.pop();

        if (topElement != this.mappings.get(c)) {
          return false;
        }
      } 
      else {
        stack.push(c);
      }
    }

    return stack.isEmpty();
  }
}