Valid Parentheses leetcode

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.

用stack实现, 左括号都push到stack中 注意返回时候是要确保stack为空才能true

时间O(n)

 
public class Solution {
    public boolean isValid(String s) {
        if (s == null || s.length() == 0) {
            return false;
        }
        Stack<Character> stack = new Stack<Character>();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(' ||s.charAt(i) == '[' || s.charAt(i) == '{') {
                stack.push(s.charAt(i));
            } else if (s.charAt(i) == ')' || s.charAt(i) == ']' || s.charAt(i) == '}') {
                if (stack.size() == 0) {
                    return false;
                }
                char c = stack.pop();
                if (s.charAt(i) == ')') {
                    if (c != '(') {
                        return false;
                    }
                } else if (s.charAt(i) == ']') {
                    if (c != '[') {
                        return false;
                    }
                } else {
                    if (c != '{') {
                        return false;
                    }
                }
            }
        }
        return stack.size() == 0; //防止有stack没清空
    }
}
//
public class Solution {
    public boolean isValid(String s) {
        HashMap<Character, Character> map = new HashMap<Character, Character>();
        map.put('(', ')');
        map.put('{', '}');
        map.put('[', ']');
        Stack<Character> stack = new Stack<Character>();
        for (int i = 0; i < s.length(); i++) {
            if (map.containsKey(s.charAt(i))) {
                stack.push(s.charAt(i));
            } else {
                if (stack.isEmpty() || map.get(stack.pop()) != s.charAt(i)) {
                    return false;
                }
            }
        }
        return stack.isEmpty();
    }
}