LeetCode – Valid Parentheses (Java)

1805 단어 LeetCode
Problem:
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.
Thoughts about This Problem
Character is not a frequently used class, so need to know how to use it.
Java Solution
public static 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++) {
		char curr = s.charAt(i);
 
		if (map.keySet().contains(curr)) {
			stack.push(curr);
		} else if (map.values().contains(curr)) {
			if (!stack.empty() && map.get(stack.peek()) == curr) {
				stack.pop();
			} else {
				return false;
			}
		}
	}
 
	return stack.empty();
}

Simplified Java Solution
Almost identical, but convert string to char array at the beginning.
public static boolean isValid(String s) {
	char[] charArray = s.toCharArray();
 
	HashMap<Character, Character> map = new HashMap<Character, Character>();
	map.put('(', ')');
	map.put('[', ']');
	map.put('{', '}');
 
	Stack<Character> stack = new Stack<Character>();
 
	for (Character c : charArray) {
		if (map.keySet().contains(c)) {
			stack.push(c);
		} else if (map.values().contains(c)) {
			if (!stack.isEmpty() && map.get(stack.peek()) == c) {
				stack.pop();
			} else {
				return false;
			}
		}
	}
	return stack.isEmpty();
}

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