leetcode刷题（20）——判断括号的有效性

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

leetcode中不需要导包 

class Solution {
    public boolean isValid(String s) {
        Stack<Character> sk = new Stack<>();
        Character c;
        for(int i=0;i<s.length();i++){
            c = s.charAt(i);  //字符串常用函数
            if(c=='('||c=='['||c=='{'){sk.push(c);}
            //注意!sk.isEmpty()是为了防止字符串后面多出括号')'出现无效
            else if(c==')'&&!sk.isEmpty()&&sk.peek()=='('){sk.pop();}  
            else if(c==']'&&!sk.isEmpty()&&sk.peek()=='['){sk.pop();}
            else if(c=='}'&&!sk.isEmpty()&&sk.peek()=='{'){sk.pop();}
            else{return false;}
        }
        return sk.isEmpty();  //栈刚好为空则正好匹配，不为空则证明字符串后面少括号
    }
}