---------------🎈🎈题目链接🎈🎈-------------------

解法1 栈的操作（先进后出）

新建栈：Stack<Character> mystack = new Stack<>();
遍历字符串：for(int i = 0; i < s.length() ; i++){char ch = s.charAt(i)}
判断栈是否为空：mystack.isEmpty()
向栈内加入元素（栈顶）：mystack.push()
从栈内移除元素（栈顶）：mystack.pop()
获取栈顶元素：mystack.peek()

时间复杂度O(N)
空间复杂度O(N)

class Solution {public boolean isValid(String s) {if(s.length() == 0) return true;Stack<Character> mystack = new Stack<>();for(int i = 0; i < s.length(); i++){char ch = s.charAt(i);if(mystack.isEmpty()){mystack.push(ch);continue;}else if((mystack.peek()=='(' && ch==')') || (mystack.peek()=='[' && ch==']') || (mystack.peek()=='{' && ch=='}')){mystack.pop();continue;}else{mystack.push(ch);}}if(mystack.isEmpty()) return true;else return false;}
}