leetcode_Valid Parentheses

描述：

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

思路：

很简单的字符串匹配问题，直接用栈就可以实现，为使程序更加高效，当要匹配的字符种类比较多的时候可以考虑用HashMap来保存匹配对

代码：

用if else来比较匹配对

public  boolean isValid(String s) {        if(s==null||s.length()==0)            return true;        Stack<Character> st=new Stack<Character>();        char ch;        int len=s.length();        for(int i=0;i<len;i++)        {            ch=s.charAt(i);            if(!st.empty())            {                if(ch==')'&&st.peek()=='(')                {                    st.pop();                }else if(ch==']'&&st.peek()=='[')                {                    st.pop();                }                else if(ch=='}'&&st.peek()=='{')                {                    st.pop();                }else                     st.push(ch);            }else                st.push(ch);                    }        if(!st.empty())            return false;        return true;            }

用HashMap来保存匹配对

public  boolean isValid(String s) {if (s == null || s.length() == 0)return true;Stack<Character> st = new Stack<Character>();Map<Character, Character> map = new HashMap<Character, Character>();map.put('(', ')');map.put('{', '}');map.put('[', ']');int len=s.length();for (int i = 0; i < len; i++) {if (!st.empty()&&map.containsKey(st.peek())&&s.charAt(i) == map.get(st.peek())) {st.pop();}elsest.push(s.charAt(i));}return st.empty();}