20. Valid Parentheses

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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.

public class Solution {    public boolean isValid(String s) {       if(s.length()%2==1 || s.length()==0) return false;        Map<Character, Character> map = new HashMap<Character, Character>();//hashmap存括号对，key存括号左半部分，value存括号有半部分        map.put('(', ')');        map.put('[', ']');        map.put('{', '}');        Deque<Character> stack = new LinkedList<>();//堆栈存括号对的左半部分        for (int i = 0; i < s.length(); i++) {            char curr = s.charAt(i);             //判断字符，如果存在于map中key中，则是左半部分，存于堆栈中。如果存在于map中的value中，则是右半部分，和堆栈中的栈顶元素比较，若相等，括号配对成功；否则返回括号配对失败。            if (map.keySet().contains(curr)) {                stack.addFirst(curr);            } else if (map.values().contains(curr)) {                if (!stack.isEmpty() && map.get(stack.getFirst()) == curr) {                    stack.removeFirst();                } else {                    return false;                }            }        }        return stack.isEmpty();    }}

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