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


正常的代碼

class Solution {
    public boolean isValid(String s) {
        Stack<Character> st=new Stack<Character>();
        for(char c:s.toCharArray()){
            if(c==‘(‘
 
||c==‘[‘||c==‘{‘)
                st.push(c);
            else if(c==‘}‘&&!st.empty()&&st.peek()==‘{‘)
                st.pop();
            else if(c==‘]‘&&!st.empty()&&st.peek()==‘[‘)
                st.pop();
            else if(c==‘)‘&&!st.empty()&&st.peek()==‘
 
(‘)
                st.pop();
            else 
                return false;
        }
        return st.empty();
    }
}

很巧妙的方法

class Solution {
    public boolean isValid(String s) {
        Stack<Character> st=new Stack<Character>();
        for(char c:s.toCharArray()){
            if(c==‘(‘)
                st.push(‘)‘);
            else if(c==‘[‘)
                st.push(‘]‘);
            else if(c==‘{‘)
                st.push(‘}‘);
            else if(st.empty()||st.pop()!=c)
                return false;
        }
        if(st.empty())
            return true;
        else
            return false;
    }
}

Leetcod--20. Valid Parentheses(極簡潔的括號匹配)